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AP Calculus BC : Inflection Points

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Example Questions

Example Question #31 : Derivatives

$\begin{align*}&\text{Find any points of inflection for the function:}\\&ln(-2x) - x - 8x^{2} - 8x^{3} + x^{5} - 2\end{align*}$

Possible Answers:

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

-1.334,-0.471,1.697

-5.335,-0.157,0.424

-0.267,-0.094,8.487

Correct answer:

-1.334,-0.471,1.697

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&\frac{1}{x}- 16x - 24x^{2} + 5x^{4} - 1\\&\text{And our second is:}\\&20x^{3} -\frac{ 1}{x^{2}}- 48x - 16\\&\text{For a reminder of derivative rules:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-1.334,-0.471,1.697,0.054-0.210i,0.054+0.210i\\&\text{We find real unique roots at the points }x=-1.334,-0.471,1.697\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

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Example Question #32 : Derivatives

$\begin{align*}&\text{Calculate any and all points of inflection for: }\\&f(x)=6x +\frac{ 1}{2^x}- 2x^{2} - 3x^{3} + 8\end{align*}$

Possible Answers:

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

-0.192

-0.096

-1.342

Correct answer:

-0.192

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&6 -\frac{ ln(2)}{2^x}- 9x^{2} - 4x\\&\text{And our second is:}\\&ln(\frac{2)^{2}}{2^x}- 18x - 4\\&\text{For a reminder of derivative rules:}\\&d[b^{ax}]=ab^{ax}ln(b)dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-0.192\\&\text{We find one real unique root at the point }x=-0.192\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

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Example Question #33 : Derivatives

$\begin{align*}&\text{Find any points of inflection for the function:}\\&ln(4x) - 4x + 10x^{2} - 7x^{3} + 5x^{4} + 6\end{align*}$

Possible Answers:

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

-0.046,0.555

-0.031,0.056

-0.183,0.278

Correct answer:

-0.183,0.278

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&20x +\frac{ 1}{x}- 21x^{2} + 20x^{3} - 4\\&\text{And our second is:}\\&60x^{2} -\frac{ 1}{x^{2}}- 42x + 20\\&\text{For a reminder of derivative rules:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-0.183,0.278,0.303+0.485i,0.303-0.485i\\&\text{We find real unique roots at the points }x=-0.183,0.278\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at every point the concavity of the function changes.}\end{align*}$

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Example Question #34 : Derivatives

$\begin{align*}&\text{Find any points of inflection for the function:}\\&3x + e^{(-x)} + 3x^{2} + 10x^{3} + 7x^{4} - 4\end{align*}$

Possible Answers:

-0.303

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

-0.152

-0.022

Correct answer:

-0.152

Explanation:

$\begin{align*}&\text{A point of inflection of a function is where its concavity changes}\\&\text{sign. To find the concavity of a function, we'll wish to first}\\&\text{find its second derivative. For the function we're given, our}\\&\text{first derivative is :}\\&6x - e^{(-x)} + 30x^{2} + 28x^{3} + 3\\&\text{And our second is:}\\&60x + e^{(-x)} + 84x^{2} + 6\\&\text{For a reminder of derivative rules:}\\&d[e^{ax}]=ae^{ax}dx\\&\text{Now, if a point of inflection exists, it's where this function equals zero with real roots.}\\&\text{Solving for x with a calculator yields:}\\&x=-0.152\\&\text{We find one real unique root at the point }x=-0.152\\&\text{Now to determine if a point is one of inflection, plug in values on either side of it,}\\&\text{and see if the sign changes. Say }\pm0.005\text{ its value.}\\&\text{With this check we find that at this point the concavity of the function changes.}\end{align*}$

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Example Question #35 : Derivatives

Find the points of inflection on the function's domain:

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

Possible Answers:

$x=\frac{1}{12}$

$x=-\frac{1}{12}$

None of the other answers

There are no points of inflection

x=-3

Correct answer:

$x=-\frac{1}{12}$

Explanation:

The points of inflection are the points at which a function's second derivative changes in sign.

To start, we must find the function's second derivative:

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

$f''=6x+\frac{1}{2}$

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

$6x+\frac{1}{2}=0$

$x=-\frac{1}{12}$

Using this value as a bound, we create intervals on which to evaluate the sign of the second derivative:

$(-\infty, -\frac{1}{12})$ . $(-\frac{1}{12}, \infty)$

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, a point of inflection exists at $x=-\frac{1}{12}$ , because the second derivative did change sign at this point.

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Example Question #37 : Derivatives

At what value of does the function

f(x)=x^2-x^3

have an inflection point?

Possible Answers:

$\frac{1}{3}$

$\frac{2}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

To find inflection points (if there are any), we compute the second derivative of f(x) and attempt to set it to . In this case:

f'(x)=2x-3x^2

f''(x)=2-6x=0

$x=\frac{1}{3}$ .

Note: To confirm this is an inflection point (where the concavity of the function actually changes sign ––not simply touches without changing sign––) we can take the derivative yet again:

f'''(x)=-6 ,

so of course, at our point of interest where f''(x)=0 ,

$f'''\left (\frac{1}{3} \right )=-6$ .

Since f'''(x) is nonzero and negative, we know that the second derivative changes sign from positive to negative, and so at x=1/3 the function changes from concave up, to concave down.

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AP Calculus BC : Inflection Points

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #31 : Derivatives

Example Question #32 : Derivatives

Example Question #33 : Derivatives

Example Question #34 : Derivatives

Example Question #35 : Derivatives

Example Question #37 : Derivatives

All AP Calculus BC Resources

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