AP Calculus BC : Derivatives

Study concepts, example questions & explanations for AP Calculus BC

varsity tutors app store varsity tutors android store

Example Questions

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

Find all the points of inflection of

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

Possible Answers:

\(\displaystyle \\ t_1=\frac{\sqrt{3}}{3} \\ \\ t_2=0\)

There are no points of inflection.

\(\displaystyle \\ t_1=\frac{\sqrt{3}}{3} \\ \\ t_2=-\frac{\sqrt{3}}{3}\)

\(\displaystyle \\ t_1=0 \\ \\ t_2=-\frac{\sqrt{3}}{3}\)

Correct answer:

\(\displaystyle \\ t_1=\frac{\sqrt{3}}{3} \\ \\ t_2=-\frac{\sqrt{3}}{3}\)

Explanation:

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

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

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

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

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

\(\displaystyle 12t^2-4=0\).

Now we can use the quadratic equation.

Recall that the quadratic equation is

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

where a,b,c refer to the coefficients of the equation  \(\displaystyle at^2+bt+c=0\).

 

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

 

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

\(\displaystyle t=\frac{\pm\sqrt{192}}{24}=\frac{\pm8\sqrt{3}}{24}=\frac{\pm\sqrt{3}}{3}\)

Thus the possible points of infection are

\(\displaystyle t_1=\frac{\sqrt{3}}{3}\)

\(\displaystyle t_2=-\frac{\sqrt{3}}{3}\).

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 \(\displaystyle t_1\) lets plug in \(\displaystyle x=0, x=1\).

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

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

Therefore \(\displaystyle t_1\) is an inflection point. 

Now lets check \(\displaystyle t_2\) with \(\displaystyle x=0, x=-1\).

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

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

Therefore \(\displaystyle t_2\) is also an inflection point. 

Example Question #1 : Points Of Inflection

Find all the points of infection of

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

Possible Answers:

There are no points of inflection.

\(\displaystyle \\ t_1=0\)

\(\displaystyle \\ \\ t_1=\frac{3\sqrt{10}}{10} \\ \\ t_2=-\frac{3\sqrt{10}}{10}\)

\(\displaystyle \\ t_1=0 \\ \\ t_2=\frac{3\sqrt{10}}{10} \\ \\ t_3=-\frac{3\sqrt{10}}{10}\)

Correct answer:

\(\displaystyle \\ t_1=0 \\ \\ t_2=\frac{3\sqrt{10}}{10} \\ \\ t_3=-\frac{3\sqrt{10}}{10}\)

Explanation:

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

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

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

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

Now lets factor \(\displaystyle f''(t)\).

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

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

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

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

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

Recall that the quadratic equation is

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

 

\(\displaystyle at^2+bt+c=0\).

 

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

 

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

\(\displaystyle t=\frac{\pm\sqrt{1440}}{40}=\frac{\pm12\sqrt{10}}{40}=\frac{\pm3\sqrt{10}}{10}\)

Thus the other 2 points of infection are

\(\displaystyle t_2=\frac{3\sqrt{10}}{10}\)

\(\displaystyle t_3=-\frac{3\sqrt{10}}{10}\)

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 \(\displaystyle x=-1, x=-\frac{1}{2}, x=\frac{1}{2}, x=1\) 

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

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

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

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

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

Example Question #2 : Inflection Points

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

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

Possible Answers:

\(\displaystyle (\pi,12\pi)\)

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

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

\(\displaystyle (12\pi,\pi)\)

Correct answer:

\(\displaystyle (\pi,12\pi)\)

Explanation:

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

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

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:

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

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

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

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

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

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

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

So our POI is:

\(\displaystyle (\pi,12\pi)\)

 

Example Question #1 : Inflection Points

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

Possible Answers:

\(\displaystyle f(x)=e^{-x}\)

\(\displaystyle f(x)=e^{x}\)

\(\displaystyle f(x)=\ln(x)\)

\(\displaystyle f(x)=x\)

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

Correct answer:

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

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. 

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

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

 

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

 

\(\displaystyle \frac{d^2}{dx^{2}}[\ln(x)]=\frac{-1}{x^{2}}\neq0\), so no inflection points here.

 

This leaves us with 

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

 

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

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

So, \(\displaystyle f''(x)=0\) when \(\displaystyle 1-100e^{-x}=0\). So 

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

Example Question #11 : Second Derivatives

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

Possible Answers:

\(\displaystyle -0.889\)

\(\displaystyle -4.444\)

\(\displaystyle \text{There are no points of inflection.}\)

\(\displaystyle -1.778\)

Correct answer:

\(\displaystyle -0.889\)

Explanation:

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

Example Question #12 : Second Derivatives

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

Possible Answers:

\(\displaystyle \text{There are no points of inflection.}\)

\(\displaystyle 3.031\)

\(\displaystyle 1.010\)

\(\displaystyle 5.051\)

Correct answer:

\(\displaystyle 1.010\)

Explanation:

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

Example Question #13 : Second Derivatives

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

Possible Answers:

\(\displaystyle \text{There are no points of inflection.}\)

\(\displaystyle -0.637,0.262\)

\(\displaystyle -3.820,1.570\)

\(\displaystyle -1.273,0.785\)

Correct answer:

\(\displaystyle -0.637,0.262\)

Explanation:

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

Example Question #11 : Inflection Points

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

Possible Answers:

\(\displaystyle -0.537,-0.177\)

\(\displaystyle \text{There are no points of inflection.}\)

\(\displaystyle -3.221,-1.242\)

\(\displaystyle -2.148,-0.710\)

Correct answer:

\(\displaystyle -0.537,-0.177\)

Explanation:

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

Example Question #11 : Second Derivatives

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

Possible Answers:

\(\displaystyle -1.151,0.651\)

\(\displaystyle \text{There are no points of inflection.}\)

\(\displaystyle -0.576,0.326\)

\(\displaystyle -3.454,1.628\)

Correct answer:

\(\displaystyle -0.576,0.326\)

Explanation:

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

Example Question #11 : Second Derivatives

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

Possible Answers:

\(\displaystyle \text{There are no points of inflection.}\)

\(\displaystyle 0.000,0.414,23.171\)

\(\displaystyle 0.000,1.036,28.964\)

\(\displaystyle 0.000,0.207,5.793\)

Correct answer:

\(\displaystyle 0.000,0.207,5.793\)

Explanation:

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

Learning Tools by Varsity Tutors