Relationship between the concavity of ƒ and the sign of ƒ'' - AP Calculus AB

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Question

What is the concavity of at x=1?

Answer

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.

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Question

Find the second derivative.

Answer

We need to find the second derivative. To do this we will use power rule twice. We will take the first derivative by power rule, then the second derivative by power rule once again. Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. $n(x^{n-1})$

Therefore, the first derivative is and the second derivative is by power rule.

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Question

$\begin{align}&\text{For the point }x=4\text{, find the concavity of the function}\&y=-20tan(3x)\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\&\text{Our first derivative is thus:}\&- 60tan(3x)^{2} - 60\&\text{And our second is:}\&-120tan(3x)\cdot (3tan(3x)^{2} + 3)\&\text{Now if we plug in our x-value we find}\&-120tan(3(4))\cdot (3tan(3(4))^{2} + 3)=321.46\&\text{The function is concave upward.}\end{align}$

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Question

$\begin{align}&\text{For the point }x=-4\text{, find the concavity of the function}\&y=13cos(17x)\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[cos(ax)]=-asin(ax)dx\&d[sin(ax)]=acos(ax)dx\&\text{Our first derivative is thus:}\&-221sin(17x)\&\text{And our second is:}\&-3757cos(17x)\&\text{Now if we plug in our x-value we find}\&-3757cos(17(-4))=-1653.62\&\text{The function is concave downward.}\end{align}$

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Question

$\begin{align}&\text{Find the concavity of the function }\&f(x)=-12tan(13x)\&\text{When }x=4\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[tan(ax)]=\frac{adx}{cos^2(ax)}=(atan^2(ax)+a)dx\&\text{Our first derivative is thus:}\&- 156tan(13x)^{2} - 156\&\text{And our second is:}\&-312tan(13x)\cdot (13tan(13x)^{2} + 13)\&\text{Now if we plug in our x-value we find}\&-312tan(13(4))\cdot (13tan(13(4))^{2} + 13)=9.24\cdot10^{5}\&\text{The function is concave upward.}\end{align}$

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Question

$\begin{align}&\text{Find the concavity of the function }\&f(x)=17e^{(6x)}\&\text{When }x=3\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[e^{ax}]=ae^{ax}dx\&\text{Our first derivative is thus:}\&102e^{(6x)}\&\text{And our second is:}\&612e^{(6x)}\&\text{Now if we plug in our x-value we find}\&612e^{(6(3))}=4.02\cdot10^{10}\&\text{The function is concave upward.}\end{align}$

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

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

Answer

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

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Question

$\begin{align}&\text{Find the concavity of the function }\&f(x)=-20ln(11x)\&\text{When }x=5\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[ln(ax)]=\frac{dx}{x}\&d[x^a]=ax^{a-1}dx\&\text{Our first derivative is thus:}\&-\frac{20}{x}\&\text{And our second is:}\&\frac{20}{x^{2}}\&\text{Now if we plug in our x-value we find}\&\frac{20}{(5)^{2}}=0.80\&\text{The function is concave upward.}\end{align}$

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Question

$\begin{align}&\text{For the point }x=9\text{, find the concavity of the function}\&y=8sin(20x)\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[sin(ax)]=acos(ax)dx\&d[cos(ax)]=-asin(ax)dx\&\text{Our first derivative is thus:}\&160cos(20x)\&\text{And our second is:}\&-3200sin(20x)\&\text{Now if we plug in our x-value we find}\&-3200sin(20(9))=2563.69\&\text{The function is concave upward.}\end{align}$

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Question

$\begin{align}&\text{For the point }x=-2\text{, find the concavity of the function}\&y=2cos(8x)\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[cos(ax)]=-asin(ax)dx\&d[sin(ax)]=acos(ax)dx\&\text{Our first derivative is thus:}\&-16sin(8x)\&\text{And our second is:}\&-128cos(8x)\&\text{Now if we plug in our x-value we find}\&-128cos(8(-2))=122.58\&\text{The function is concave upward.}\end{align}$

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Question

$\begin{align}&\text{For the point }x=-10\text{, find the concavity of the function}\&y=\frac{14}{x^{16}}\end{align}$

Answer

$\begin{align}&\text{To find the concavity of a function, we'll wish to first find}\&\text{its second derivative, so let's do it in steps. For the function}\&\text{we're given,}\text{ we'll wish to use the following:}\&d[x^a]=ax^{a-1}dx\&\text{Our first derivative is thus:}\&-\frac{224}{x^{17}}\&\text{And our second is:}\&\frac{3808}{x^{18}}\&\text{Now if we plug in our x-value we find}\&\frac{3808}{(-10)^{18}}=3.81\cdot10^{-15}\&\text{The function is concave upward.}\end{align}$

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Question

Tell whether f is concave up or concave down when . How do you know?

Answer

Tell whether f is concave up or concave down when . How do you know?

To test for concavity, we need to find the second derivative.

In this case, we can use the power rule to do all our differentiation.

Power rule:

$\frac{d}{dx}x^n=(n)x^{n-1}$

We will use this on each term in order to find our first and then second derivative.

For each term, we will decrease the exponent by 1, and then multiply by the original exponent.

Do the same thing to f'(x) to get our final function.

Now, to test for concavity, we will plug in our x value (5), and look at the sign of the result.

SO, our second derivative is positive, very positive.

This means that f(x) is concave up at x=5.

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Question

Determine the intervals on which the function is concave down:

$f=\frac{x^3}{2}+2x$

Answer

To determine the intervals on which the function is concave down, we must determine the intervals on which the function's second derivative is negative.

First, we must find the second derivative of the function:

$f'=\frac{3}{2}x^2+2$

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 must find the values at which the second derivative is equal to zero:

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

$(-\infty, 0), (0, \infty)$

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 negative, while on the second interval, the second derivative is positive. Therefore, the function is concave down on the first interval, $(-\infty, 0)$ .

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Question

Determine the intervals on which the function is concave up:

Answer

To determine the intervals on which the function is concave up, we must determine the intervals on which the function's second derivative is positive.

First, we must find the second derivative of the function:

The derivatives were found using the following rules:

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

The second derivative is a constant and is negative, which means that the function is concave down on its domain, never concave up.

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Question

$f(x) = x^{3}+ 4x^{2}+ 8x- 17$

A relative minimum of the graph of can be located at (nearest hundredth):

Answer

At a relative minimum of the graph , it will hold that and .

First, find . Using the sum rule,

$f(x) = x^{3}+ 4x^{2}+ 8x- 17$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}( x^{3}+ 4x^{2}+ 8x- 17)$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}( x^{3} )+ \frac{\mathrm{d} }{\mathrm{d} x}( 4x^{2} )+\frac{\mathrm{d} }{\mathrm{d} x}( 8x )-\frac{\mathrm{d} }{\mathrm{d} x}( 17)$

Apply the Constant Multiple and Power Rules:

$f'(x) =3 x^{2} + 4 \cdot 2x +8x -0$

$f'(x) =3 x^{2} + 8x +8x$

Set this equal to 0:

$3 x^{2} + 8x +8x =0$

$x= \frac{-8 \pm \sqrt{8^{2}-4(3)(8)}}{2(3)} = \frac{-8 \pm \sqrt{-32}}{6}$

Both zeroes of are imaginary. Since has no real zeroes, it follows that the graph of has no relative extrema - maximum or minimum - on the coordinate plane.

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Question

Determine the intervals on which the function is concave down:

$f=\frac{x^5}{20}+\frac{x^4}{12}+\frac{x^3}{6}+2x^2$

Answer

To determine the intervals on which the function is concave down, we must determine the intervals on which the function's second derivative is negative.

First, we must find the second derivative of the function,

$f'=\frac{x^4}{4}+\frac{x^3}{3}-2x^2+4x$

which 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 find the values at which the second derivative is equal to zero:

Note that we factored by grouping to solve for x.

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

$(-\infty, -2), (-2, -1), (-1, 2), (2, \infty)$

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 negative, on the second interval, the second derivative is positive, on the third interval, the second derivative is negative, and on the fourth interval, the second derivative is positive. The function is concave down on the intervals where the second derivative is negative, $(-\infty, -2)\cup(-1, 2)$ .

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Question

Determine the intervals on which the function is concave up:

$f=\frac{x^4}{12}-\frac{5x^3}{6}-3x^2$

Answer

To determine the intervals on which the function is concave up, we must determine the intervals on which the function's second derivative is positive.

First, we must find the second derivative of the function,

$f'=\frac{x^3}{3}-\frac{5x^2}{2}-6x$

which 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 find the values at which the second derivative is equal to zero:

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

$(-\infty, -1), (-1, 6), (6, \infty)$

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, the second derivative is negative, and on the third interval, the second derivative is positive. So, the function is concave up on the intervals $(-\infty, -1)\cup(6, \infty)$

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