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AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : How To Graph Functions Of Points Of Inflection

Find all the points of inflection of

f(t)=t^4-2t^2+5t+100

Possible Answers:

$\\ t_1=\frac{\sqrt{3}}{3} \\ \\ t_2=0$

There are no points of inflection.

$\\ t_1=\frac{\sqrt{3}}{3} \\ \\ t_2=-\frac{\sqrt{3}}{3}$

$\\ t_1=0 \\ \\ t_2=-\frac{\sqrt{3}}{3}$

Correct answer:

$\\ t_1=\frac{\sqrt{3}}{3} \\ \\ t_2=-\frac{\sqrt{3}}{3}$

Explanation:

In order to find the points of inflection, we need to find f''(t) using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

f(t)=t^4-2t^2+5t+100

f'(t)=4t^3-4t+5

f''(t)=12t^2-4

Now to find the points of inflection, we need to set f''(t)=0 .

12t^2-4=0 .

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 at^2+bt+c=0 .

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 t_1 lets plug in x=0, x=1 .

f''(0)=12(0)^2-4=-4

f''(1)=12(1)^2-4=8

Therefore t_1 is an inflection point.

Now lets check t_2 with x=0, x=-1 .

f''(0)=12(0)^2-4=-4

f''(-1)=12(-1)^2-4=8

Therefore t_2 is also an inflection point.

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Example Question #1 : Points Of Inflection

Find all the points of infection of

f(t)=t^5-3t^3+2t+100 .

Possible Answers:

There are no points of inflection.

$\\ t_1=0$

$\\ \\ t_1=\frac{3\sqrt{10}}{10} \\ \\ t_2=-\frac{3\sqrt{10}}{10}$

$\\ t_1=0 \\ \\ t_2=\frac{3\sqrt{10}}{10} \\ \\ t_3=-\frac{3\sqrt{10}}{10}$

Correct answer:

$\\ t_1=0 \\ \\ t_2=\frac{3\sqrt{10}}{10} \\ \\ t_3=-\frac{3\sqrt{10}}{10}$

Explanation:

In order to find the points of inflection, we need to find f''(t) using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

f(t)=t^5-3t^3+2t+100

f'(t)=5t^4-9t^2+2

f''(t)=20t^3-18t

Now lets factor f''(t) .

f''(t)=t(20t^2-18)

Now to find the points of inflection, we need to set f''(t)=0 .

t(20t^2-18)=0 .

From this equation, we already know one of the point of inflection, t_1=0 .

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

at^2+bt+c=0 .

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''(-1)=20(-1)^3-18(-1)=-2

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

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

f''(1)=20(1)^3-18(1)=2

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

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Example Question #11 : Second Derivatives

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

f(x)=5sin(x)+12x

Possible Answers:

$\left(6\pi,\frac{\pi}{2}\right)$

$\left(\frac{\pi}{2},6\pi\right)$

$(\pi,12\pi)$

$(12\pi,\pi)$

Correct answer:

$(\pi,12\pi)$

Explanation:

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:

f'(x)=5cos(x)+12

f''(x)=-5sin(x)

So, where on the given interval does f''(x)=0 ?

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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Example Question #281 : Ap Calculus Bc

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

Possible Answers:

$f(x)=\frac{10}{1+100e^{-x}}$

$f(x)=\ln(x)$

$f(x)=e^{x}$

$f(x)=e^{-x}$

f(x)=x

Correct answer:

$f(x)=\frac{10}{1+100e^{-x}}$

Explanation:

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.

$\frac{d^2}{dx^{2}}[e^{x}]=e^{x}\neq0$ , there are no real values of for which this equals zero, so no inflection points.

$\frac{d^2}{dx^{2}}[e^{-x}]=e^{-x}\neq0$ , same story here.

$\frac{d^2}{dx^{2}}[\ln(x)]=\frac{-1}{x^{2}}\neq0$ , 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, f''(x)=0 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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Example Question #11 : Second Derivatives

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

Possible Answers:

-0.889

-4.444

$\text{There are no points of inflection.}$

-1.778

Correct answer:

-0.889

Explanation:

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

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Example Question #12 : Second Derivatives

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

Possible Answers:

$\text{There are no points of inflection.}$

3.031

1.010

5.051

Correct answer:

1.010

Explanation:

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

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Example Question #13 : Second Derivatives

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

Possible Answers:

$\text{There are no points of inflection.}$

-0.637,0.262

-3.820,1.570

-1.273,0.785

Correct answer:

-0.637,0.262

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative, so let's do it in steps. For the}\\&\text{function we're given, our first derivative is :}\\&9x^{2} - 8x + 16x^{3} + 8\\&\text{And our second is:}\\&18x + 48x^{2} - 8\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x yields:}\\&x=-0.637,0.262\\&\text{We find real unique roots at the points }x=-0.637,0.262\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

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Example Question #11 : Inflection Points

$\begin{align*}&\text{Calculate any and all points of inflection for: }\\&f(x)=1 - 10x^{3} - 7x^{4} - 4x^{2}\end{align*}$

Possible Answers:

-0.537,-0.177

$\text{There are no points of inflection.}$

-3.221,-1.242

-2.148,-0.710

Correct answer:

-0.537,-0.177

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative, so let's do it in steps. For the}\\&\text{function we're given, our first derivative is :}\\&- 8x - 30x^{2} - 28x^{3}\\&\text{And our second is:}\\&- 60x - 84x^{2} - 8\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x yields:}\\&x=-0.537,-0.177\\&\text{We find real unique roots at the points }x=-0.537,-0.177\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

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Example Question #11 : Second Derivatives

$\begin{align*}&\text{Determine what, if any exist, the points of inflection are for the function}\\&y=9x^{2} - x - 4x^{3} - 8x^{4} - 5\end{align*}$

Possible Answers:

-1.151,0.651

$\text{There are no points of inflection.}$

-0.576,0.326

-3.454,1.628

Correct answer:

-0.576,0.326

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative, so let's do it in steps. For the}\\&\text{function we're given, our first derivative is :}\\&18x - 12x^{2} - 32x^{3} - 1\\&\text{And our second is:}\\&18 - 96x^{2} - 24x\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x yields:}\\&x=-0.576,0.326\\&\text{We find real unique roots at the points }x=-0.576,0.326\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

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Example Question #11 : Second Derivatives

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

Possible Answers:

$\text{There are no points of inflection.}$

0.000,0.414,23.171

0.000,1.036,28.964

0.000,0.207,5.793

Correct answer:

0.000,0.207,5.793

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative, so let's do it in steps. For the}\\&\text{function we're given, our first derivative is :}\\&12x^{2} - 40x^{3} + 5x^{4} - 3\\&\text{And our second is:}\\&24x - 120x^{2} + 20x^{3}\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x yields:}\\&x=0.000,0.207,5.793\\&\text{We find real unique roots at the points }x=0.000,0.207,5.793\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

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AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #4 : How To Graph Functions Of Points Of Inflection

Example Question #1 : Points Of Inflection

Example Question #11 : Second Derivatives

Example Question #281 : Ap Calculus Bc

Example Question #11 : Second Derivatives

Example Question #12 : Second Derivatives

Example Question #13 : Second Derivatives

Example Question #11 : Inflection Points

Example Question #11 : Second Derivatives

Example Question #11 : Second Derivatives

All AP Calculus BC Resources

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