Second Derivatives - AP Calculus AB

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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.

What is the concavity of at x=1?

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First, find the double derivative of the function given.

You should get .

Next, plug in x=1 to get .

The double derivative is a positive number, which means the concavity is Concave Up.

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}$=x$

$\begin{align}&$\text{Determine the concavity of the $function}$\&-9e^{(-6x)}$\&$\text{At the point }$x=-8\end{align}$

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$\begin{align}&$\text{To find the concavity of a function, we'll wish to first find}$\&$\text{its second derivative, so let's do it in steps. For the function}$\&$\text{we're given,}$$\text{ we'll wish to use the $following:}$\&d[e^{ax}$$]=ae^{ax}$dx\&$\text{Our first derivative is $thus:}$\&54e^{(-6x)}$\&$\text{And our second $is:}$\&-324e^{(-6x)}$\&$\text{Now if we plug in our x-value we $find}$\&-324e^{(-6(-8))}$$=-2.27\cdot10^{23}$\&$\text{The function is concave downward.}$\end{align}$

$\begin{align}&$\text{Determine the concavity of the $function}$\&$\frac{6}{x^{10}$$}\&$\text{At the point }$x=-1\end{align}$

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$\begin{align}&$\text{To find the concavity of a function, we'll wish to first find}$\&$\text{its second derivative, so let's do it in steps. For the function}$\&$\text{we're given,}$$\text{ we'll wish to use the $following:}$\&d[x^a$$]=ax^{a-1}$dx\&$\text{Our first derivative is $thus:}$\&-$\frac{60}{x^{11}$$}\&$\text{And our second $is:}$\&$\frac{660}{x^{12}$$}\&$\text{Now if we plug in our x-value we $find}$\&$\frac{660}{(-1)^{12}$$}=660.00\&$\text{The function is concave upward.}$\end{align}$

$\begin{align}&$\text{Determine the concavity of the function}$\&-cos(15x)\&$\text{At the point }$x=6\end{align}$

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$\begin{align}&$\text{To find the concavity of a function, we'll wish to first find}$\&$\text{its second derivative, so let's do it in steps. For the function}$\&$\text{we're given,}$$\text{ we'll wish to use the following:}$\&d[cos(ax)]=-asin(ax)dx\&d[sin(ax)]=acos(ax)dx\&$\text{Our first derivative is thus:}$\&15sin(15x)\&$\text{And our second is:}$\&225cos(15x)\&$\text{Now if we plug in our x-value we find}$\&225cos(15(6))=-100.82\&$\text{The function is concave downward.}$\end{align}$

$\begin{align}&$\text{Determine the concavity of the $function}$\&$\frac{19}{5^{(5x)}$$}\&$\text{At the point }$x=-5\end{align}$

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$\begin{align}&$\text{To find the concavity of a function, we'll wish to first find}$\&$\text{its second derivative, so let's do it in steps. For the function}$\&$\text{we're given,}$$\text{ we'll wish to use the $following:}$\&d[b^{ax}$$]=ab^{ax}$ln(b)sx\&$\text{Our first derivative is $thus:}$\&-$\frac{(95ln(5))}{5^{(5x)}$$}\&$\text{And our second $is:}$\&$\frac{(475ln(5)^{2}$$$)}{5^{(5x)}$}\&$\text{Now if we plug in our x-value we $find}$\&$\frac{(475ln(5)^{2}$$$)}{5^{(5(-5))}$$}=3.67\cdot10^{20}$\&$\text{The function is concave upward.}$\end{align}$

Find the points of inflection of

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We will find the points of inflection by setting our second derivative to zero.

Take the derivative again,

then set this equal to zero and reverse foil,

$0=x-1 \ or \ 0=x+4$

$x=1 \or \ x=-4$

These are our points of inflection.

h(x)=\frac{f+g}{f}$

If

and ,

then find .

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We see the answer is when we use the product rule.

h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}$$}

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}$$} =-2

On what intervals is the function both concave up and decreasing?

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The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

Solving for the zero's, we see hits zero at and . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

Thus, we can see that the derivative is only negative on the interval .

Repeating the process for the second derivative,

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is

On a closed interval, the function is decreasing. What can we say about and on these intervals?

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If is decreasing, then its derivative is negative. The derivative of is , so this is telling us that is negative.

For to be decreasing, would have to be negative, which we don't know.

being negative has nothing to do with its slope.

For to be decreasing, its derivative would need to be negative, or, alternatively would have to be concave down, which we don't know.

Thus, the only correct answer is that is negative.

If is a twice-differentiable function, and , find the values of the inflection point(s) of on the interval $(0,2\pi)$ .

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To find the inflection points of , we need to find (which lucky for us, is already given!) set it equal to , and solve for .

. Start

$0=\sin(x)$ . Divide by . We can do this, because is never equal to .

On the unit circle, the values $0, \pi, 2\pi$ cause $\sin(x)=0$ , but only $\pi$ is inside our interval $(0,2\pi)$ . so $x=\pi$ is the only value to consider here.

To prove that $x = \pi$ is actually part of a point of inflection, we have to test an value on the left and the right of $\pi$ , and substitute them into and test their signs.

$f''(\pi/2) = $\sin(\pi/2)e^{(\pi/2)}$ = $e^{(\pi/2)}$ >0$

$f''(3\pi/2) = $\sin(3\pi/2)e^{(3\pi/2)}$ = $-e^{(3\pi/2)}$<0$ .

Hence $x=\pi$ , is the coordinate of an inflection point of on the interval $(0,2\pi)$ .

At what value(s) does the following function have an inflection point?

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Inflection points of a function occur when the second derivative equals zero. Therefore, we simply need to take two derivatives of our function, and solve.

Therefore, the two values that makes this function go to zero are

Determine the points of inflection for the following function:

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To determine the points of inflection, we must find the value at which the second derivative of the function changes in sign.

First, we find the second derivative:

The derivatives were found using the following rules:

$\frac{\mathrm{d}$ }{\mathrm{d} x} $x^n$$=nx^{n-1}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} ax=a$

Next, we find the values at which the second derivative of the function is equal to zero:

Using the critical value, we now create intervals over which to evaluate the sign of the second derivative:

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, while on the second interval, the second derivative is positive. The second derivative changed sign , so there exists the point of inflection for the function.

Determine the points of inflection for the following function:

$f=3\cos(2x), 0 \leq x \leq \pi$

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To determine the points of inflection, we must find the value at which the second derivative of the function changes in sign.

First, we find the second derivative:

$f'=-6\sin(2x)$

$f''=-12\cos(2x)$

The derivatives were found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(x)=\cos(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} ax=a$

Next, we find the values at which the second derivative of the function is equal to zero:

$-12\cos(2x)=0$

$x=\frac{\pi}{4}$, $\frac{3\pi}{4}$$

Note that the x values we find are limited by the interval given in the problem statement.

Using the critical value, we now create intervals over which to evaluate the sign of the second derivative:

$(0, $\frac{\pi}{4}$), ($\frac{\pi}{4}$, $\frac{3\pi}{4}$), ($\frac{3\pi}{4}$, \pi)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is positive, on the second interval, the second derivative is negative, and on the third interval, the second derivative is positive. Thus, two points of inflection exist where the second derivative changed in sign, $x=\frac{\pi}{4}$, $\frac{3\pi}{4}$$ .

For the function y, at which point(s) does the rate of _change_in the slope shift from decreasing to increasing.

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For this question we want to find the points at which the second derivative of the function changes sign, i.e. . These are points at which the slope of the first derivative changes signs.

$x=\pm $\sqrt{\frac{2}{3}$}$

Notice that there are two x-values at which the second derivative equals zero.

$y=2(\pm $$\sqrt{\frac{2}{3}$})^4$-8(\pm $$\sqrt{\frac{2}{3}$})^2$$

$y=2($\frac{4}{9}$)-8($\frac{2}{3}$)$

$y=\frac{8}{9}$-$\frac{16}{3}$$

$y=-$\frac{40}{9}$$

We are interested only in the point where the slope changes from decreasing to increasing, however. This means that the answer is at $x=+$\sqrt{\frac{2}{3}$}$ .

x-values less than $+\sqrt$\frac{2}{3}$$ (and greater than $-\sqrt$\frac{2}{3}$$ ) yield negative y-values, and x-values greater than $+\sqrt$\frac{2}{3}$$ yield positive y-values. In other words, $(\sqrt$\frac{2}{3}$, -$\frac{40}{9}$)$ is the only point at which the second derivative changes from negative to positive, which represents the slope (first derivative) changing from decreasing to increasing.

A computer program generates the following codes when it finds points of inflection:

$\alpha$ , when it finds no points of inflection;
$\beta$ , when it finds one point of inflection;
$\gamma$ , when it finds two points of inflection;
$\delta$ , when it finds three or more points of inflection.

What code will the computer generate for the following function:

$f=4 \cos(x), 0 \leq x \leq 2\pi$

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To find what code the computer generates, we must figure out the points of inflection for the function, which are the points at which the second derivative switches sign.

First, we find the second derivative:

$f'=-4\sin(x)$

$f''(x)=-4\cos(x)$

The derivatives were found using the following rules:

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(x)=\cos(x)$

Next, we must find the x values - on the given interval - for which the second derivative is equal to zero:

$-4\cos(x)=0$

$x=\frac{\pi}{2}$, $\frac{3\pi}{2}$$

Using these values as endpoints, we create the intervals on which we evaluate the sign of the second derivative:

$(0, $\frac{\pi}{2}$), ($\frac{\pi}{2}$, $\frac{3\pi}{2}$), ($\frac{3\pi}{2}$, 2\pi)$

Note that at the endpoints of each interval, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, on the second interval, the second derivative is positive, and on the third interval, the second derivative is negative. The sign of the second derivative changed twice, so points of inflection exist at $x=\frac{\pi}{2}$, $\frac{3\pi}{2}$$ .

The computer will generate the code for two points of inflection as $\gamma$ .

Determine the points of inflection of the following function:

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To determine the points of inflection for the function, we must points at which the second derivative of the function changes sign.

First, we find the second derivative of the function:

and was found using the following rules:

$\frac{\mathrm{d}$ }{\mathrm{d} x} $x^n$$=nx^{n-1}$ , $$\frac{\mathrm{d}$ }{\mathrm{d} x} ax=a$

Next, we must the values of x for which the second derivative is equal to zero:

Using the critical values, we now create intervals on which to evaluate the sign of the second derivative:

$(-\infty, -3) ,(-3, 1), (1, \infty)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is positive, on the second interval, it is negative, and on the third interval, it is positive. Thus, two points of inflection exist for the function, at because the second derivative changed sign at each of these locations.