High School Math : Applications of Derivatives

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #41 : Second Derivatives

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

Give the interval(s) on which \displaystyle g is decreasing.

Possible Answers:

\displaystyle (-\infty ,\infty)

\displaystyle (3,\infty)

\displaystyle (-\infty ,3)

\displaystyle (2,4)

\displaystyle (-\infty ,2) \cup (4,\infty)

Correct answer:

\displaystyle (2,4)

Explanation:

\displaystyle g is decreasing on those intervals at which \displaystyle g'(x)< 0.

 

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

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

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

 

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

\displaystyle 3x^{2} -18x+24 =0

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

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

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

 

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

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

 

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

 

\displaystyle (-\infty ,2)

Choose \displaystyle x = 0

\displaystyle 3x^{2} -18x+24 =3\cdot 0^{2} -18\cdot 0+24 =24 > 0

\displaystyle g increases on this interval.

 

\displaystyle (2,4)

Choose \displaystyle x = 3

\displaystyle 3x^{2} -18x+24 =3\cdot 3^{2} -18\cdot 3+24 =27 -54 +24 = -3 < 0

\displaystyle g decreases on this interval.

 

\displaystyle (4,\infty)

Choose \displaystyle x = 5

\displaystyle 3x^{2} -18x+24 =3\cdot 5^{2} -18\cdot 5+24 = 75 -90 +24 = 9 > 0

\displaystyle g increases on this interval.

 

The answer is that \displaystyle g decreases on \displaystyle (2,4).

Example Question #2 : Applications Of Derivatives

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

Give the interval(s) on which \displaystyle f is increasing.

Possible Answers:

\displaystyle (-\infty ,2) \cup (5,\infty)

\displaystyle \left ( 3 \frac{1}{2}, \infty \right )

\displaystyle (-\infty ,\infty)

\displaystyle \left ( -\infty , 3 \frac{1}{2}\right )

\displaystyle (2,5)

Correct answer:

\displaystyle (-\infty ,2) \cup (5,\infty)

Explanation:

\displaystyle f is increasing on those intervals at which \displaystyle f'(x)>0.

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

 

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

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

 

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

\displaystyle 3x^{2} - 21 x +30 = 0

\displaystyle 3(x^{2} - 7 x +10 )= 0

\displaystyle 3(x-2)(x-5)= 0

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

 

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

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

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

 

\displaystyle (-\infty ,2)

Choose \displaystyle x = 0

\displaystyle 3x^{2} - 21 x +30 = 3 \cdot 0^{2} - 21 \cdot 0 +30 = 30 > 0

\displaystyle f increases on this interval.

 

\displaystyle (2,5)

Choose \displaystyle x = 3

\displaystyle 3x^{2} - 21 x +30 = 3 \cdot 3^{2} - 21 \cdot 3 +30 = 27 -63 +30 = -6 < 0

\displaystyle f decreases on this interval.

 

\displaystyle (5,\infty)

Choose \displaystyle x = 6

\displaystyle 3x^{2} - 21 x +30 = 3 \cdot 6^{2} - 21 \cdot 6 +30 = 108 -126 + 30=12 > 0

\displaystyle f increases on this interval.

 

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

Example Question #1 : Finding Regions Of Increasing And Decreasing Value

At what point does \displaystyle -8x^2+15 shift from increasing to decreasing?

Possible Answers:

It does not shift from increasing to decreasing

\displaystyle x=0

\displaystyle x=-16

\displaystyle x=-8

\displaystyle x=-\frac{1}{16}

Correct answer:

\displaystyle x=0

Explanation:

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 \displaystyle 15 as \displaystyle 15x^0 since anything to the zero power is one.

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

Notice that \displaystyle (0*15x^{0-1})=0 since anything times zero is zero.

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

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

If we were to graph \displaystyle y=-16x, would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

\displaystyle y=-16x

\displaystyle 0=-16x

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

\displaystyle 0=x

Example Question #1 : Applications Of Derivatives

At what point does \displaystyle 4x^2+\frac{2}{3}x shift from decreasing to increasing?

Possible Answers:

\displaystyle -\frac{1}{24}=x

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

\displaystyle 0=x

\displaystyle \frac{1}{8}=x

\displaystyle -12=x

Correct answer:

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

Explanation:

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.

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

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

Remember that anything to the zero power is one.

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

\displaystyle \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 \displaystyle 0=8x+\frac{2}{3}.

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

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

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

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

Example Question #2 : Finding Regions Of Increasing And Decreasing Value

At what value of \displaystyle x does \displaystyle 5x^2+12x shift from decreasing to increasing?

Possible Answers:

It does not shift from decreasing to increasing

\displaystyle x=\frac{5}{6}

\displaystyle x=-10

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

\displaystyle x=-12

Correct answer:

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

Explanation:

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.

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})

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

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

Anything to the zero power is one.

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

\displaystyle \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.

\displaystyle y=10x+12

\displaystyle 0=10x+12

\displaystyle -12=10x

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

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

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

 

Example Question #1232 : Ap Calculus Ab

At the point \displaystyle x=2, is the function \displaystyle -8x^2+15 increasing or decreasing, concave or convex?

Possible Answers:

The function is undefined at that point

Decreasing, concave

Decreasing, convex

Increasing, concave

Increasing, convex

Correct answer:

Decreasing, convex

Explanation:

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 \displaystyle 15 as \displaystyle 15x^0 since anything to the zero power is one.

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

Notice that \displaystyle (0*15x^{0-1})=0 since anything times zero is zero.

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

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

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

\displaystyle y=-16x

\displaystyle y=-16(2)

\displaystyle y=-32

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 \displaystyle -16x as our expression.

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

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

\displaystyle \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 \displaystyle x; 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.

Example Question #2 : Finding Regions Of Concavity And Convexity

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

Possible Answers:

\displaystyle \left (\frac{-1}{2}, \frac{1}{2} \right )

\displaystyle (-\infty ,0)

\displaystyle (0,\infty )

\displaystyle (1 ,e^2)

\displaystyle (-\infty ,\infty )

Correct answer:

\displaystyle (0,\infty )

Explanation:

This question asks us to examine the concavity of the function \displaystyle 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 \displaystyle f(x)=g(x)\cdot h(x), then \displaystyle f'(x)=g'(x)\cdot h(x)+h'(x)\cdot g(x). In this particular problem, let \displaystyle g(x)= x and \displaystyle h(x)=e^{x^2}. Applying the Product rule, we get

\displaystyle 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

\displaystyle 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 \displaystyle e^{x^2}, we will need to invoke the Chain Rule. According to the Chain Rule, the derivative of a function in the form \displaystyle f(x)=g(h(x)) is given by \displaystyle f'(x)=g'(h(x))\cdot h'(x). In finding the derivative of \displaystyle e^{x^2}, we will let \displaystyle g(x)=e^x and \displaystyle h(x)=x^2.

\displaystyle \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. 

 

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

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

To summarize, the first derivative of the funciton \displaystyle f(x)= x \cdot e^{x^2} is \displaystyle f'(x)=e^{x^2}(2x^2+1).

 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.

\displaystyle 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 }

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

\displaystyle =e^{x^2}\cdot(4x^3+6x)

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

\displaystyle f''(x)=e^{x^2}(4x^3+6x) > 0

In order to solve this inequality, we can divide both sides by \displaystyle e^{x^2}. Notice that \displaystyle e^{x^2} is always positive (because e raised to any power will be positive); this means that when we divide both sides of the inequality by \displaystyle e^{x^2}, 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 \displaystyle e^{x^2} gives us

\displaystyle 4x^3+6x>0

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

\displaystyle 2x(2x^2+3)>0

Notice now that the expression \displaystyle 2x^2 + 3 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 \displaystyle 2x^2 + 3 without having to change the direction of the sign. This leaves us with the inequality

\displaystyle 2x >0, which clearly only holds when \displaystyle x > 0.

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

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

 

 

Example Question #1234 : Ap Calculus Ab

At the point \displaystyle x=-5, is \displaystyle 5x^2+12x increasing or decreasing, and is it concave or convex?

Possible Answers:

Decreasing, convex

Increasing, convex

The graph is undefined at point \displaystyle x=-5

Increasing, concave

Decreasing, concave

Correct answer:

Decreasing, convex

Explanation:

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.

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})

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

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

Anything to the zero power is one.

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

\displaystyle \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.

\displaystyle y=10x+12

\displaystyle y=10(-5)+12

\displaystyle y=-50+12

\displaystyle y=-38

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 \displaystyle 10x+12 as our expression.

We're going to treat \displaystyle 12 as \displaystyle 12x^0.

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x+12)=(1*10x^{1-1})+(0*12x^{0-1})

Notice that \displaystyle (0*12x^{0-1})=0 since anything times zero is zero.

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

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

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

\displaystyle \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.

Example Question #1233 : Ap Calculus Ab

When \displaystyle x=3, what is the concavity of the graph of \displaystyle 2x^4+\frac{1}{2}x?

Possible Answers:

There is insufficient data to solve.

Decreasing, convex

Increasing, concave

Decreasing, concave

Increasing, convex

Correct answer:

Increasing, convex

Explanation:

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:

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

Simplify:

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

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

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

\displaystyle \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, \displaystyle x=3, to see if the result is positive (i.e. increasing) or negative (i.e. decreasing).

\displaystyle y=8(3)^{3}+\frac{1}{2}

\displaystyle y=216.5

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, \displaystyle x=3, and check if it is positive or negative.

We're going to treat \displaystyle \frac{1}{2} as \displaystyle \frac{1}{2}x^0 since anything to the zero power is equal to one.

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

Notice that \displaystyle (0*\frac{1}{2}x^{0-1})=0 since anything times zero is zero.

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

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

Plug in our given value:

\displaystyle y=24(3)^2

\displaystyle y=216

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.

Example Question #2 : Finding Regions Of Concavity And Convexity

At the point where \displaystyle x=5, is \displaystyle 4x^2+\frac{2}{3}x increasing or decreasing, and is it concave up or down?

Possible Answers:

Decreasing, concave up

Increasing, concave down

Decreasing, concave down

There is no concavity at that point.

Increasing, concave up

Correct answer:

Increasing, concave up

Explanation:

To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at \displaystyle x=5, 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.

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

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

Remember that anything to the zero power is one.

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

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

Plug in our given value.

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

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

\displaystyle y=40+\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 \displaystyle 8x+\frac{2}{3} as our expression.

For this problem, we're going to say that \displaystyle \frac{2}{3}=\frac{2}{3}x^0 since, as stated before, anything to the zero power is one.

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

Notice that \displaystyle (0*\frac{2}{3}x^{0-1})=0 as anything times zero is zero.

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

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

\displaystyle \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.

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