Second Derivatives - 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

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

Give the interval(s) on which is increasing.

Answer

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:

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

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

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

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

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

increases on this interval.

Choose

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

decreases on this interval.

$(5,\infty)$

Choose

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

increases on this interval.

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

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Question

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

Give the interval(s) on which is decreasing.

Answer

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:

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

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

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

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

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

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

increases on this interval.

Choose

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

decreases on this interval.

$(4,\infty)$

Choose

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

increases on this interval.

The answer is that decreases on .

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Question

At what value of does shift from decreasing to increasing?

Answer

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.

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Question

At what point does shift from increasing to decreasing?

Answer

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 $(015x^{0-1})=0$ 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.

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

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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

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

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

Find the points of inflection of $f(x)=\frac{1}{12}x^4+\frac{1}{2}x^3-2x^2+6x-2$

Answer

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

$f'(x)=\frac{1}{3}x^3+\frac{3}{2}x^2-4x+6$

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.

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Question

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

If

$f^{'}(x)=2$

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Answer

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

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

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

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

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Question

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

Answer

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

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Question

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

Answer

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.

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Question

If is a twice-differentiable function, and $f''(x)=\sin(x)e^x$ , find the values of the inflection point(s) of on the interval $(0,2\pi)$ .

Answer

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

$0=\sin(x)e^x$ . 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 $(0,2\pi)$ .

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Question

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

Answer

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

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Question

Determine the points of inflection for the following function:

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

Answer

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.

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