Second Derivatives - AP Calculus BC

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Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

$=-2x\ln(x)+-x^2(\frac{1}{x})$

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

$=-(2\ln x +1)+-x\cdot\frac{2}{x}=-2\ln x-1-2=-2\ln x-3$

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

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Question

Determine the intervals on which the function is concave up:

$f=4\sin(\frac{x}{2})$

on the interval $0\leq x\leq 4\pi$

Answer

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

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

$f'=2\cos(\frac{x}{2})$

$f''=-\sin(\frac{x}{2})$

The derivatives were found using the following rules:

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

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

$-\sin(\frac{x}{2})=0$

$x=0, 2\pi, 4\pi$

We now use these values as bounds for the intervals on which we check the sign of the second derivative:

$(0, 2\pi), (2\pi, 4\pi)$

Note that 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. Thus, the function is concave up on the second interval, $(2\pi, 4\pi)$ .

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Question

Tell whether g(w) is concave up or concave down when . How do you know?

Answer

Tell whether g(w) is concave up or concave down when . How do you know?

To test for concavity, we need to know the sign of our function's second derivative at the given value.

Let's begin by finding the first derivative. To do so, we need to recall the following:

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

$2)\frac{d}{dx}e^x=e^x$

Those two rules tell us how to derive functions made up of polynomials and exponential terms.

Apply the rules to get the following.

$g'(w)=-(3)13t^{3-1}-(2)12t-6e^t$

$g'(w)=-39t^{2}-24t-6e^t$

Now, do derive again to get the 2nd derivative.

$g'(w)=-39t^{2}-24t-6e^t$

Now, we are almost there, but we need to find g"(-3)

$g''(-3)=-78(-3)-24-6e^{-3}=234-24-0.2987224\approx209.7013$

So, our second derivative is positive when w=-3. This means our original function is concave up at this point.

Concave up, because

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Question

At which of the following values of is the function

concave down?

Answer

The function is concave down wherever , so we compute and see where it is negative. We have:

(a parabola, opening upwards)

To find where is negative, we first find its zeros by setting :

,

so when or , and we conclude that is negative ( is concave down) between them. That is, . The only answer choice completely inside this interval (not outside or at the endpoints) is

$x=\frac{7}{4}$ .

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Question

Over which of the following intervals is the following function concave upward?

$f(x)=-x^2\ln(x)$

Answer

In order to determine the concavity of a function, we need to examine its second derivative.

First, we will take the derivative of the function by applying the product rule.

$f'(x)=(\ln (x))\cdot\frac{\mathrm{d} [-x^2]}{\mathrm{d} x}+-x^2\cdot\frac{\mathrm{d} [\ln x]}{\mathrm{d} x}$

$=-2x\ln(x)+-x^2(\frac{1}{x})$

$=-2x\ln(x)-x=-x(2\ln x+1)$

We will then take the derivative of the first derivative. Again, we will employ the product rule.

$f''(x)=(2\ln(x)+1)\cdot\frac{\mathrm{d} }{\mathrm{d} x}(-x)+-x\cdot\frac{\mathrm{d} }{\mathrm{d} x}(2\ln(x)+1)$

$=-(2\ln x +1)+-x\cdot\frac{2}{x}=-2\ln x-1-2=-2\ln x-3$

A funciton is concave upward whenever its second derivative is positive. Thus, we must find the interval over which f''(x) is positive.

$-2\ln x -3 > 0$

$\ln x < -\frac{3}{2}$

This will only be true when $x<e^{\frac{-3}{2}}$ . However, keep in mind that the domain of f(x) will include only positive values of x, because lnx is only defined for x > 0.

The answer is { ${}{x:0<x<e^{\frac{-3}{2}}}$ }.

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Question

Determine the intervals on which the function is concave up:

$f=4\sin(\frac{x}{2})$

on the interval $0\leq x\leq 4\pi$

Answer

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

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

$f'=2\cos(\frac{x}{2})$

$f''=-\sin(\frac{x}{2})$

The derivatives were found using the following rules:

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

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

$-\sin(\frac{x}{2})=0$

$x=0, 2\pi, 4\pi$

We now use these values as bounds for the intervals on which we check the sign of the second derivative:

$(0, 2\pi), (2\pi, 4\pi)$

Note that 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. Thus, the function is concave up on the second interval, $(2\pi, 4\pi)$ .

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Question

Tell whether g(w) is concave up or concave down when . How do you know?

Answer

Tell whether g(w) is concave up or concave down when . How do you know?

To test for concavity, we need to know the sign of our function's second derivative at the given value.

Let's begin by finding the first derivative. To do so, we need to recall the following:

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

$2)\frac{d}{dx}e^x=e^x$

Those two rules tell us how to derive functions made up of polynomials and exponential terms.

Apply the rules to get the following.

$g'(w)=-(3)13t^{3-1}-(2)12t-6e^t$

$g'(w)=-39t^{2}-24t-6e^t$

Now, do derive again to get the 2nd derivative.

$g'(w)=-39t^{2}-24t-6e^t$

Now, we are almost there, but we need to find g"(-3)

$g''(-3)=-78(-3)-24-6e^{-3}=234-24-0.2987224\approx209.7013$

So, our second derivative is positive when w=-3. This means our original function is concave up at this point.

Concave up, because

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Question

At which of the following values of is the function

concave down?

Answer

The function is concave down wherever , so we compute and see where it is negative. We have:

(a parabola, opening upwards)

To find where is negative, we first find its zeros by setting :

,

so when or , and we conclude that is negative ( is concave down) between them. That is, . The only answer choice completely inside this interval (not outside or at the endpoints) is

$x=\frac{7}{4}$ .

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Question

What type of point is $\small (0,0)$ in ?

Answer

Even though the first derivative ( $\small \small 3x^2$ ) is $\small 0$ at $\small x=0$ , there is no max or min because the function is increasing on both sides (derivative is positive on both sides). However, the function is changing its concavity (from negative to positive) at $\small x=0$ . We can tell because the second derivative ( $\small 6x$ ) is also $\small 0$ at $\small x=0$ , and it's going from negative to positive. Hole and asymptote are irrelevant. When the second derivative changes signs around a specific point we call this an inflection point. An inflection point describes a point that changes the concavity of a function.

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Question

Find the inflection point(s) of .

Answer

Inflection points can only occur when the second derivative is zero or undefined. Here we have

.

Therefore possible inflection points occur at and $x=\frac{3}{2}$ . However, to have an inflection point we must check that the sign of the second derivative is different on each side of the point. Here we have

.

Hence, both are inflection points

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Question

Below is the graph of . How many inflection points does have?

Answer

Possible inflection points occur when . This occurs at three values, . However, to be an inflection point the sign of must be different on either side of the critical value. Hence, only are critical points.

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Question

Find the point(s) of inflection for the function .

Answer

A point of inflection is found where the graph (or image) of a function changes concavity. To find this algebraically, we want to find where the second derivative of the function changes sign, from negative to positive, or vice-versa. So, we find the second derivative of the given function

The first derivative using the power rule

$y=x^n \rightarrow y'=nx^{n-1}$ is,

and the seconds derivative is

We then find where this second derivative equals . when .

We then look to see if the second derivative changes signs at this point. Both graphically and algebraically, we can see that the function does indeed change sign at, and only at, , so this is our inflection point.

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Question

What are the coordinates of the points of inflection for the graph

$y=\frac{x^4}{12}+x^3+4x^2-8x+5?$

Answer

Infelction points are the points of a graph where the concavity of the graph changes. The inflection points of a graph are found by taking the double derivative of the graph equation, setting it equal to zero, then solving for .

To take the derivative of this equation, we must use the power rule,

$\frac{d}{dx}x^n=nx^{n-1}$ .

We also must remember that the derivative of an constant is 0.

After taking the first derivative of the graph equation using the power rule, the equation becomes

$y^{'}=\frac{x^3}{3}+3x^2+8x-8$ .

In this problem the double derivative of the graph equation comes out to $y^{''}=x^2+6x+8$ , factoring this equation out it becomes $y^{''}=(x+4)(x+2)$ .

Solving for when the equation is set equal to zero, the inflection points are located at .

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Question

Find all the points of inflection of

Answer

In order to find the points of inflection, we need to find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Now to find the points of inflection, we need to set .

.

Now we can use the quadratic equation.

Recall that the quadratic equation is

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ ,

where a,b,c refer to the coefficients of the equation .

In this case, a=12, b=0, c=-4.

$t=\frac{-0\pm\sqrt{0^2-4(12)(-4)}}{(2)(12)}$

$t=\frac{\pm\sqrt{192}}{24}=\frac{\pm8\sqrt{3}}{24}=\frac{\pm\sqrt{3}}{3}$

Thus the possible points of infection are

$t_1=\frac{\sqrt{3}}{3}$

$t_2=-\frac{\sqrt{3}}{3}$ .

Now to check if or which are inflection points we need to plug in a value higher and lower than each point. If there is a sign change then the point is an inflection point.

To check lets plug in .

Therefore is an inflection point.

Now lets check with .

Therefore is also an inflection point.

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Question

Find all the points of infection of

.

Answer

In order to find the points of inflection, we need to find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Now lets factor .

Now to find the points of inflection, we need to set .

.

From this equation, we already know one of the point of inflection, .

To figure out the rest of the points of inflection we can use the quadratic equation.

Recall that the quadratic equation is

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , where a,b,c refer to the coefficients of the equation

.

In this case, a=20, b=0, c=-18.

$t=\frac{-0\pm\sqrt{0^2-4(20)(-18)}}{(2)(20)}$

$t=\frac{\pm\sqrt{1440}}{40}=\frac{\pm12\sqrt{10}}{40}=\frac{\pm3\sqrt{10}}{10}$

Thus the other 2 points of infection are

$t_2=\frac{3\sqrt{10}}{10}$

$t_3=-\frac{3\sqrt{10}}{10}$

To verify that they are all inflection points we need to plug in values higher and lower than each value and see if the sign changes.

Lets plug in $x=-1, x=-\frac{1}{2}, x=\frac{1}{2}, x=1$

$f''\left(-\frac{1}{2}\right)=20\left(-\frac{1}{2} \right )^3-18\left(-\frac{1}{2} \right )=6.5$

$f''\left(\frac{1}{2}\right)=20\left(\frac{1}{2} \right )^3-18\left(\frac{1}{2} \right )=-6.5$

Since there is a sign change at each point, all are points of inflection.

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Question

Which of the following is a point of inflection of on the interval $\left[\frac{\pi}{2},\pi\right]$ ?

Answer

Which of the following is a point of inflection of f(x) on the interval $\left[\frac{\pi}{2},\pi\right]$ ?

To find points of inflection, we need to know where the second derivative of the function is equal to zero. So, find the second derivative:

So, where on the given interval does ?

Well, we know from our unit circle that $sin(\pi)=0$ ,

So we would have a point of inflection at $x=\pi$ , but we still need to find the y-coordinate of our POI. find this by finding $f(\pi)$

$f(\pi)=5sin(\pi)+12(\pi)=12\pi$

So our POI is:

$(\pi,12\pi)$

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Question

Which of the following is true about the twice-differentiable function above?

Answer

Since the function is increasing at , , and since is below the x-axis, .

Furthermore, there exists an inflection point at , where the concavity of function changes.

Thus at , .

Therefore, the correct answer is .

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Question

Which of the following functions has an inflection point where concavity changes?

Answer

For a graph to have an inflection point, the second derivative must be equal to zero. We also want the concavity to change at that point.

$\frac{d^2}{dx^{2}}[x]=0$ , for all real numbers, but this is a line and has no concavity associated with it, so not this one.

$\frac{d^2}{dx^{2}}[e^{x}]=e^{x} eq0$ , there are no real values of for which this equals zero, so no inflection points.

$\frac{d^2}{dx^{2}}[e^{-x}]=e^{-x} eq0$ , same story here.

$\frac{d^2}{dx^{2}}[\ln(x)]=\frac{-1}{x^{2}} eq0$ , so no inflection points here.

This leaves us with

$f(x)=\frac{10}{1+100e^{-x}}$ , whose derivatives are a bit more difficult to take.

$f(x)=10(1+100e^{-x})^{-1}$ , so by the chain rule we get $f'(x)=-10(1+100e^{-x})^{-2}(-100e^{-x})=1000e^{-x}(1+100e^{-x})^{-2}$

$f''(x)=-1000e^{-x}(1+100e^{-x})^{-2}-2000e^{-x}(1+100e^{-x})^{-3}(-100e^{-x})=-1000e^{-x}(\frac{1+100e^{-x}-200e^{-x}}{(1+100e^{-x})^{3}})=-1000e^{-x}(\frac{1-100e^{-x}}{(1+100e^{-x})^{3}})$

So, when $1-100e^{-x}=0$ . So

$-100e^{-x}=-1\ e^{-x}=1/100\ \frac{1}{e^{x}}=\frac{1}{100}\ e^{x}=100 x=\ln(100)$ . This is our correct answer.

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Question

$\begin{align}&\text{Calculate any and all points of inflection for: }\&f(x)=6x - 8x^{2} - 3x^{3} - 5\end{align}$

Answer

$\begin{align}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative, so let's do it in steps. For the}\&\text{function we're given, our first derivative is :}\&6 - 9x^{2} - 16x\&\text{And our second is:}\&- 18x - 16\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x yields:}\&x=-0.88889\&\text{We find one real unique root at the point }x=-0.889\&\text{Now to determine if a point is one of inflection, check values on either side of it. Say }\pm0.005/\text{ its value.}\&\text{With this check we find that at this point the concavity of the function changes.}\end{align}$

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Question

$\begin{align}&\text{Determine what, if any exist, the points of inflection are for the function}\&y=6x^{5} - 7x^{2} - 8x^{3} - 5x^{4} - 10x + 1\end{align}$

Answer

$\begin{align}&\text{A point of inflection of a function is where its concavity changes}\&\text{sign. To find the concavity of a function, we'll wish to first}\&\text{find its second derivative, so let's do it in steps. For the}\&\text{function we're given, our first derivative is :}\&30x^{4} - 24x^{2} - 20x^{3} - 14x - 10\&\text{And our second is:}\&120x^{3} - 60x^{2} - 48x - 14\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\&\text{Solving for x yields:}\&x=1.0103,-0.25513+0.22449i,-0.25513-0.22449i\&\text{We find one real unique root at the point }x=1.010\&\text{Now to determine if a point is one of inflection, check values on either side of it. Say }\pm0.005/\text{ its value.}\&\text{With this check we find that at this point the concavity of the function changes.}\end{align}$

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