Applications of Derivatives - Math

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When , what is the concavity of the graph of ?

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To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=(42x^{4-1}$$)+(1$\frac{1}{2}$x^{1-1}$)$

Simplify:

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=(42x^{3}$$)+($\frac{1}{2}$x^{0}$)$

Remember that anything to the zero power is equal to one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=(8x^{3}$)+($\frac{1}{2}$1)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=8x^{3}$+\frac{1}{2}$

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, , to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, , and check if it is positive or negative.

We're going to treat $\frac{1}{2}$ as since anything to the zero power is equal to one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x^{3}$$+\frac{1}{2}$)=(38x^{3-1}$$)+(0$\frac{1}{2}$x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x^{3}$$+\frac{1}{2}$)=(3*8x^{2}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x^{3}$$+\frac{1}{2}$)=24x^{2}$

Plug in our given value:

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

At the point , is increasing or decreasing, and is it concave or convex?

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To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{0}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$$+12x^0$

Anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$+12(1)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$+12x)=10x+12$

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using as our expression.

We're going to treat as .

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(10x+12)=(110x^{1-1}$$)+(012x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(10x+12)=(1*10x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(10x+12)=(10x^{0}$)$

As stated before, anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} x}(10x+12)=10$

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

At the point , is the function increasing or decreasing, concave or convex?

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First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$+15)=-16x$

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using as our expression.

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-16x)=1-16x^{1-1}$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-16x)=-16x^{0}$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

At the point where , is increasing or decreasing, and is it concave up or down?

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To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{1}$$)+(1$\frac{2}{3}$x^{0}$)$

Remember that anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=(8x)+(1$\frac{2}{3}$(1))$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=8x+\frac{2}{3}$

Plug in our given value.

$y=8x+\frac{2}{3}$$

$y=8(5)+\frac{2}{3}$$

$y=40+\frac{2}{3}$$

$y=40\t$\frac{2}{3}$$

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use $8x+\frac{2}{3}$$ as our expression.

For this problem, we're going to say that since, as stated before, anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x+\frac{2}{3}$)=(18x^{1-1}$$)+(0$\frac{2}{3}$x^{0-1}$)$

Notice that as anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x+\frac{2}{3}$)=(18x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x+\frac{2}{3}$)=(8x^{0}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}(8x+\frac{2}{3}$)=8$

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

Let $f(x)= x \cdot $e^{x^2}$$ . What is the largest interval of x for which f(x) is concave upward?

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This question asks us to examine the concavity of the function $f(x)= x \cdot $e^{x^2}$$ . We will need to find the second derivative in order to determine where the function is concave upward and downward. Whenever its second derivative is positive, a function is concave upward.

Let us begin by finding the first derivative of f(x). We will need to use the Product Rule. According to the Product Rule, if $f(x)=g(x)\cdot h(x)$ , then $f'(x)=g'(x)\cdot h(x)+h'(x)\cdot g(x)$ . In this particular problem, let and . Applying the Product rule, we get

$f'(x)=\left ($\frac{\mathrm{d}$ }{\mathrm{d} x}[x] \right )\cdot $e^{x^2}$+\left ($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{ x^2}$] \right )\cdot x$

$f'(x)=1\cdot $e^{x^2}$+\left ($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{ x^2}$] \right )\cdot x$

In order to evaluate the derivative of , we will need to invoke the Chain Rule. According to the Chain Rule, the derivative of a function in the form is given by $f'(x)=g'(h(x))\cdot h'(x)$ . In finding the derivative of , we will let and .

$\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{x^2}$$]=e^{x^2}$\cdot $\frac{\mathrm{d}$ }{\mathrm{d} $x}[x^2$]=2x\cdot $e^{x^2}$

We can now finish finding the derivative of the original function.

$f'(x)=1\cdot $e^{x^2}$+\left($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{ x^2}$] \right)\cdot x$

To summarize, the first derivative of the funciton $f(x)= x \cdot $e^{x^2}$$ is .

We need the second derivative in order to examine the concavity of f(x), so we will differentiate one more time. Once again, we will have to use the Product Rule in conjunction with the Chain Rule.

$f''(x)=\frac{\mathrm{d}$ }{\mathrm{d} x}[f'(x)]=\left ($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{x^2}$] \right $)\cdot(2x^2$+1)+\left ($\frac{\mathrm{d}$ }{\mathrm{d} x} [ $2x^2$+1] \right ) \cdot $e^{x^2 }$$

$=2xe^ ${x^2$} \cdot $(2x^2$+1) + 4x\cdot e^ ${x^2$$}=e^{x^2}$$\cdot(2x(2x^2$+1)+4x)$

In order to find where f(x) is concave upward, we must find where f''(x) > 0.

In order to solve this inequality, we can divide both sides by . Notice that is always positive (because e raised to any power will be positive); this means that when we divide both sides of the inequality by , we won't have to flip the sign. (If we divide an inequality by a negative quantity, the sign flips.)

Dividing both sides of the inequality by gives us

When solving inequalities with polynomials, we often need to factor.

Notice now that the expression will always be positive, because the smallest value it can take on is 3, when x is equal to zero. Thus, we can safely divide both sides of the inequality by without having to change the direction of the sign. This leaves us with the inequality

, which clearly only holds when .

Thus, the second derivative of f''(x) will be positive (and f(x) will be concave up) only when . To represent this using interval notation (as the answer choices specify) we would write this as $(0,\infty )$ .

The answer is $(0,\infty )$ .

Define $f \left ( x\right ) = $x^{3}$ - $\frac{21}{2}$ $x^{2}$ +30x -6$ .

Give the interval(s) on which is increasing.

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is increasing on those intervals at which .

$f \left ( x\right ) = $x^{3}$ - $\frac{21}{2}$ $x^{2}$ +30x -6$

$f' \left ( x\right ) = $3x^{3-1}$ - 2 \cdot $\frac{21}{2}$ $x^{2-1}$ +30 -0$

$f' \left ( x\right ) = $3x^{2}$ - 21 x +30$

We need to find the values of for which $f' \left ( x\right ) = $3x^{2}$ - 21 x +30 > 0$ . To that end, we first solve the equation:

$x = 2 \textrm{ or }x = 5$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , , and $(5,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(5,\infty)$

Choose

increases on this interval.

The answer is that increases on $(-\infty ,2) \cup (5,\infty)$

Define $g(x) = $x^{3}$ $-9x^{2}$+24x+17$ .

Give the interval(s) on which is decreasing.

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is decreasing on those intervals at which .

$g(x) = $x^{3}$ $-9x^{2}$+24x+17$

$g'(x) = 3\cdot $x^{3-1}$ -2\cdot $9x^{2-1}$+24+0$

$g'(x) = $3x^{2}$ -18x+24$

We need to find the values of for which $g'(x) = $3x^{2}$ -18x+24 <0$ . To that end, we first solve the equation:

$3 \left $(x^{2}$ -6x+8 \right ) =0$

$3 \left (x-2 \right )\left (x-4 \right ) =0$

$x=2\textrm{ or } x = 4$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , , and $(4,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(4,\infty)$

Choose

increases on this interval.

The answer is that decreases on .

At what value of does shift from decreasing to increasing?

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To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{0}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$$+12x^0$

Anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$+12(1)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$+12x)=10x+12$

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

$$\frac{-12}{10}$=x$

$$\frac{-6}{5}$=x$

This is the point where the graph shifts from decreasing to increasing.

At what point does shift from increasing to decreasing?

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To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2*-8x^{1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$+15)=-16x$

If we were to graph , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

$$\frac{0}{-16}$=x$

At what point does shift from decreasing to increasing?

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To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{1}$$)+(1$\frac{2}{3}$x^{0}$)$

Remember that anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=(8x)+(1*$\frac{2}{3}$(1))$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=8x+\frac{2}{3}$

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that $0=8x+\frac{2}{3}$$ .

$0=8x+\frac{2}{3}$$

$-$\frac{2}{3}$=8x$

$$\frac{-\frac{2}{3}$}{8}=x$

$-$\frac{1}{12}$=x$

When , what is the concavity of the graph of ?

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To find the concavity, we need to look at the first and second derivatives at the given point.

To take the first derivative of this equation, use the power rule. The power rule says that we lower the exponent of each variable by one and multiply that number by the original exponent:

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=(42x^{4-1}$$)+(1$\frac{1}{2}$x^{1-1}$)$

Simplify:

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=(42x^{3}$$)+($\frac{1}{2}$x^{0}$)$

Remember that anything to the zero power is equal to one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=(8x^{3}$)+($\frac{1}{2}$1)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(2x^4$$+\frac{1}{2}$x)=8x^{3}$+\frac{1}{2}$

The first derivative tells us if the function is increasing or decreasing. Plug in the given point, , to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

Therefore the function is increasing.

To find out if the function is convex, we need to look at the second derivative evaluated at the same point, , and check if it is positive or negative.

We're going to treat $\frac{1}{2}$ as since anything to the zero power is equal to one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x^{3}$$+\frac{1}{2}$)=(38x^{3-1}$$)+(0$\frac{1}{2}$x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x^{3}$$+\frac{1}{2}$)=(3*8x^{2}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x^{3}$$+\frac{1}{2}$)=24x^{2}$

Plug in our given value:

Since the second derivative is positive, the function is convex.

Therefore, we are looking at a graph that is both increasing and convex at our given point.

At the point , is increasing or decreasing, and is it concave or convex?

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To find out if the function is increasing or decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{0}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$$+12x^0$

Anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$+12(1)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$+12x)=10x+12$

Now we plug in our given value and find out if the result is positive or negative. If it is positive, the function is increasing. If it is negative, the function is decreasing.

Therefore, the function is decreasing.

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave.

To find the second derivative, we repeat the process using as our expression.

We're going to treat as .

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(10x+12)=(110x^{1-1}$$)+(012x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(10x+12)=(1*10x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(10x+12)=(10x^{0}$)$

As stated before, anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} x}(10x+12)=10$

Since we get a positive constant, it doesn't matter where we look on the graph, as our second derivative will always be positive. That means that this graph is going to be convex at our given point.

Therefore, the function is decreasing and convex at our given point.

At the point , is the function increasing or decreasing, concave or convex?

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First, let's find out if the graph is increasing or decreasing. For that, we need the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$+15)=-16x$

Plug in our given point for . If the result is positive, the function is increasing. If the result is negative, the function is decreasing.

Our result is negative, therefore the function is decreasing.

To find the concavity, look at the second derivative. If the function is positive at our given point, it is concave. If the function is negative, it is convex.

To find the second derivative we repeat the process, but using as our expression.

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-16x)=1-16x^{1-1}$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-16x)=-16x^{0}$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}(-16x)=-16$

As you can see, our second derivative is a constant. It doesn't matter what point we plug in for ; our output will always be negative. Therefore our graph will always be convex.

Combine our two pieces of information to see that at the given point, the graph is decreasing and convex.

At the point where , is increasing or decreasing, and is it concave up or down?

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To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{1}$$)+(1$\frac{2}{3}$x^{0}$)$

Remember that anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=(8x)+(1$\frac{2}{3}$(1))$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=8x+\frac{2}{3}$

Plug in our given value.

$y=8x+\frac{2}{3}$$

$y=8(5)+\frac{2}{3}$$

$y=40+\frac{2}{3}$$

$y=40\t$\frac{2}{3}$$

Is it positive? Yes. Then it is increasing.

To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.

Repeat the process we used for the first derivative, but use $8x+\frac{2}{3}$$ as our expression.

For this problem, we're going to say that since, as stated before, anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x+\frac{2}{3}$)=(18x^{1-1}$$)+(0$\frac{2}{3}$x^{0-1}$)$

Notice that as anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x+\frac{2}{3}$)=(18x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(8x+\frac{2}{3}$)=(8x^{0}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}(8x+\frac{2}{3}$)=8$

As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.

This means that at our given point, the graph is increasing and concave up.

Let $f(x)= x \cdot $e^{x^2}$$ . What is the largest interval of x for which f(x) is concave upward?

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This question asks us to examine the concavity of the function $f(x)= x \cdot $e^{x^2}$$ . We will need to find the second derivative in order to determine where the function is concave upward and downward. Whenever its second derivative is positive, a function is concave upward.

Let us begin by finding the first derivative of f(x). We will need to use the Product Rule. According to the Product Rule, if $f(x)=g(x)\cdot h(x)$ , then $f'(x)=g'(x)\cdot h(x)+h'(x)\cdot g(x)$ . In this particular problem, let and . Applying the Product rule, we get

$f'(x)=\left ($\frac{\mathrm{d}$ }{\mathrm{d} x}[x] \right )\cdot $e^{x^2}$+\left ($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{ x^2}$] \right )\cdot x$

$f'(x)=1\cdot $e^{x^2}$+\left ($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{ x^2}$] \right )\cdot x$

In order to evaluate the derivative of , we will need to invoke the Chain Rule. According to the Chain Rule, the derivative of a function in the form is given by $f'(x)=g'(h(x))\cdot h'(x)$ . In finding the derivative of , we will let and .

$\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{x^2}$$]=e^{x^2}$\cdot $\frac{\mathrm{d}$ }{\mathrm{d} $x}[x^2$]=2x\cdot $e^{x^2}$

We can now finish finding the derivative of the original function.

$f'(x)=1\cdot $e^{x^2}$+\left($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{ x^2}$] \right)\cdot x$

To summarize, the first derivative of the funciton $f(x)= x \cdot $e^{x^2}$$ is .

We need the second derivative in order to examine the concavity of f(x), so we will differentiate one more time. Once again, we will have to use the Product Rule in conjunction with the Chain Rule.

$f''(x)=\frac{\mathrm{d}$ }{\mathrm{d} x}[f'(x)]=\left ($\frac{\mathrm{d}$ }{\mathrm{d} $x}[e^{x^2}$] \right $)\cdot(2x^2$+1)+\left ($\frac{\mathrm{d}$ }{\mathrm{d} x} [ $2x^2$+1] \right ) \cdot $e^{x^2 }$$

$=2xe^ ${x^2$} \cdot $(2x^2$+1) + 4x\cdot e^ ${x^2$$}=e^{x^2}$$\cdot(2x(2x^2$+1)+4x)$

In order to find where f(x) is concave upward, we must find where f''(x) > 0.

In order to solve this inequality, we can divide both sides by . Notice that is always positive (because e raised to any power will be positive); this means that when we divide both sides of the inequality by , we won't have to flip the sign. (If we divide an inequality by a negative quantity, the sign flips.)

Dividing both sides of the inequality by gives us

When solving inequalities with polynomials, we often need to factor.

Notice now that the expression will always be positive, because the smallest value it can take on is 3, when x is equal to zero. Thus, we can safely divide both sides of the inequality by without having to change the direction of the sign. This leaves us with the inequality

, which clearly only holds when .

Thus, the second derivative of f''(x) will be positive (and f(x) will be concave up) only when . To represent this using interval notation (as the answer choices specify) we would write this as $(0,\infty )$ .

The answer is $(0,\infty )$ .

Define $f \left ( x\right ) = $x^{3}$ - $\frac{21}{2}$ $x^{2}$ +30x -6$ .

Give the interval(s) on which is increasing.

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is increasing on those intervals at which .

$f \left ( x\right ) = $x^{3}$ - $\frac{21}{2}$ $x^{2}$ +30x -6$

$f' \left ( x\right ) = $3x^{3-1}$ - 2 \cdot $\frac{21}{2}$ $x^{2-1}$ +30 -0$

$f' \left ( x\right ) = $3x^{2}$ - 21 x +30$

We need to find the values of for which $f' \left ( x\right ) = $3x^{2}$ - 21 x +30 > 0$ . To that end, we first solve the equation:

$x = 2 \textrm{ or }x = 5$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , , and $(5,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(5,\infty)$

Choose

increases on this interval.

The answer is that increases on $(-\infty ,2) \cup (5,\infty)$

Define $g(x) = $x^{3}$ $-9x^{2}$+24x+17$ .

Give the interval(s) on which is decreasing.

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is decreasing on those intervals at which .

$g(x) = $x^{3}$ $-9x^{2}$+24x+17$

$g'(x) = 3\cdot $x^{3-1}$ -2\cdot $9x^{2-1}$+24+0$

$g'(x) = $3x^{2}$ -18x+24$

We need to find the values of for which $g'(x) = $3x^{2}$ -18x+24 <0$ . To that end, we first solve the equation:

$3 \left $(x^{2}$ -6x+8 \right ) =0$

$3 \left (x-2 \right )\left (x-4 \right ) =0$

$x=2\textrm{ or } x = 4$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , , and $(4,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(4,\infty)$

Choose

increases on this interval.

The answer is that decreases on .

At what value of does shift from decreasing to increasing?

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To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{0}$)$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$$+12x^0$

Anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=10x^1$+12(1)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$+12x)=10x+12$

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

$$\frac{-12}{10}$=x$

$$\frac{-6}{5}$=x$

This is the point where the graph shifts from decreasing to increasing.

At what point does shift from increasing to decreasing?

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To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that since anything times zero is zero.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2*-8x^{1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$+15)=-16x$

If we were to graph , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

$$\frac{0}{-16}$=x$

At what point does shift from decreasing to increasing?

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To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{1}$$)+(1$\frac{2}{3}$x^{0}$)$

Remember that anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=(8x)+(1*$\frac{2}{3}$(1))$

$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$+\frac{2}{3}$x)=8x+\frac{2}{3}$

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that $0=8x+\frac{2}{3}$$ .

$0=8x+\frac{2}{3}$$

$-$\frac{2}{3}$=8x$

$$\frac{-\frac{2}{3}$}{8}=x$

$-$\frac{1}{12}$=x$