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Precalculus : Determine Points of Inflection

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Example Questions

Example Question #1 : Determine Points Of Inflection

Determine the location of the points of inflection for the following function:

$f(x)=\frac{1}{12}x^4-\frac{1}{2}x^3-5x^2+4x+17$

Possible Answers:

x=4,x=7

x=-5,x=2

x=-2,x=5

x=-4,x=3

x=-3,x=4

Correct answer:

x=-2,x=5

Explanation:

The points of inflection of a function are the points at which its concavity changes. The concavity of a function is described by its second derivative, which will be equal to zero at the inflection points, so we'll start by finding the first derivative of the function:

$f(x)=\frac{1}{12}x^4-\frac{1}{2}x^3-5x^2+4x+17$

$f'(x)=\frac{1}{3}x^3-\frac{3}{2}x^2-10x+4$

Next we'll take the derivative one more time to get the second derivative of the original function:

$f'(x)=\frac{1}{3}x^3-\frac{3}{2}x^2-10x+4$

f''(x)=x^2-3x-10

Now that we have the second derivative of the function, we can set it equal to 0 and solve for the points of inflection:

$x^2-3x-10=0\rightarrow (x-5)(x+2)=0\rightarrow x=-2,x=5$

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Example Question #2 : Determine Points Of Inflection

Find the points of inflection of the following function:

$f(x)=\frac{1}{4}x^4-5x^3+\frac{63}{2}x^2-21$

Possible Answers:

x=-3,x=5

x=2,x=4

x=-1,x=5

x=3,x=7

x=-11,x=-7

Correct answer:

x=3,x=7

Explanation:

The points of inflection of a function are those at which its second derivative is equal to 0. First we find the second derivative of the function, then we set it equal to 0 and solve for the inflection points:

$f(x)=\frac{1}{4}x^4-5x^3+\frac{63}{2}x^2-21$

f'(x)=x^3-15x^2+63x

f''(x)=3x^2-30x+63=3(x^2-10x+21)=3(x-3)(x-7)=0

x=3,x=7

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Example Question #1 : Determine Points Of Inflection

Find the inflection points of the following function:

$f(x)=\frac{1}{6}x^4-x^3-28x^2+9$

Possible Answers:

x=-7,x=4

x=-4,x=7

$x=-3,x=-\frac{2}{3}$

x=-5,x=3

x=-3,x=5

Correct answer:

x=-4,x=7

Explanation:

$f(x)=\frac{1}{6}x^4-x^3-28x^2+9$

$f'(x)=\frac{2}{3}x^3-3x^2-56x$

f''(x)=2x^2-6x-56=2(x^2-3x-28)=2(x-7)(x+4)=0

x=-4,x=7

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Example Question #1 : Determine Points Of Inflection

Determine the points of inflection of the following function:

$f(x)=\frac{1}{12}x^4+\frac{1}{3}x^3-\frac{15}{2}x^2+17$

Possible Answers:

$x=-\frac{1}{3},x=2$

x=-2,x=4

x=-3,x=5

x=-5,x=3

x=-4,x=2

Correct answer:

x=-5,x=3

Explanation:

$f(x)=\frac{1}{12}x^4+\frac{1}{3}x^3-\frac{15}{2}x^2+17$

$f'(x)=\frac{1}{3}x^3+x^2-15x$

f''(x)=x^2+2x-15=(x+5)(x-3)=0

x=-5,x=3

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Example Question #116 : Derivatives

Determine the x-coordinate of the inflection point of the function y=x^3-4x^2+3 .

Possible Answers:

$-\frac{4}{3}$

$\frac{8}{3}$

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The point of inflection exists where the second derivative is zero.

y=x^3-4x^2+3

y'=3x^2-8x

y''=6x-8 , and we set this equal to zero.

0=6x-8

8=6x

$x=\frac{8}{6}=\frac{4}{3}$

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Example Question #117 : Derivatives

Find the x-coordinates of all points of inflection of the function y=x^4+2x^3-3x^2+1 .

Possible Answers:

$0,-\frac{3}{4}\pm\frac{\sqrt{33}}{4}$

2,-3

$-\frac{1}{3}\pm\frac{\sqrt{2}}{3}$

$-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$

There are no points of inflection

Correct answer:

$-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$

Explanation:

We set the second derivative of the function equal to zero to find the x-coordinates of any points of inflection.

y=x^4+2x^3-3x^2+1

y'=4x^3+6x^2-6x

y''=12x^2+12x-6

0=12x^2+12x-6

0=6(2x^2+2x-1)

$0\neq 6$

0=2x^2+2x-1 , and the quadratic formula yields

$x=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}$ .

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Example Question #118 : Derivatives

Determine the x-coordinate(s) of the point(s) of inflection of the function $y=\frac{1}{6}x^4-\frac{2}{3}x^3-15x^2+12x+1$ .

Possible Answers:

x=-5.6,8.2

x=0

There are no points of inflection.

x=-3

x=-3,5

Correct answer:

x=-3,5

Explanation:

Any points of inflection that exist will be found where the second derivative is equal to zero.

$y=\frac{1}{6}x^4-\frac{2}{3}x^3-15x^2+12x+1$

$y'=\frac{2}{3}x^3-2x^2-30x+12$

y''=2x^2-4x-30

0=2x^2-4x-30

0=2(x^2-2x-15) .

Since $0\neq2$ , we can focus on 0=x^2-2x-15 . Thus

0=(x+3)(x-5) , and x=-3,5 .

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Example Question #1 : Determine Points Of Inflection

Find the x-coordinate(s) of the point(s) of inflection of $y=\frac{1}{6}x^4-x^2-3x+1$ .

Possible Answers:

There are no inflection points.

$x=0,\pm1$

x=-1

$x=\pm1$

x=1

Correct answer:

$x=\pm1$

Explanation:

The inflection points, if they exist, will occur where the second derivative is zero.

$y=\frac{1}{6}x^4-x^2-3x+1$

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

y''=2x^2-2

0=2x^2-2

0=2(x^2-1)

0=x^2-1

$x=\pm1$

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Example Question #2 : Determine Points Of Inflection

Find the point(s) of inflection of the function y=4x^3-3x+2 .

Possible Answers:

x=0

There is no point of inflection.

(2,28)

(0,2)

x=2

Correct answer:

(0,2)

Explanation:

The point of inflection will exist where the second derivative equals zero.

y=4x^3-3x+2

y'=12x^2-3

y''=24x

0=24x

x=0 .

Now we need the y-coordinate of the point.

y=4(0)^3-3(0)+2=2

Thus the inflection point is at (0,2) .

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Example Question #1 : Determine Points Of Inflection

Find the point of inflection of the function y=2x^3-3x^2+4 .

Possible Answers:

(1,3)

(0,4)

(2,0)

(-0.9108,0)

$\left(\frac{1}{2},\frac{7}{2}\right)$

Correct answer:

$\left(\frac{1}{2},\frac{7}{2}\right)$

Explanation:

To find the x-coordinate of the point of inflection, we set the second derivative of the function equal to zero.

y=2x^3-3x^2+4

y'=6x^2-6x

y''=12x-6

0=12x-6

6=12x

$x=\frac{6}{12}=\frac{1}{2}$ .

To find the y-coordinate of the point, we plug the x-coordinate back into the original function.

$y=2\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)^2+4$

$y=2\left(\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4$

$y=\frac{7}{2}$

The point is then $\left(\frac{1}{2},\frac{7}{2}\right)$ .

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Precalculus : Determine Points of Inflection

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Determine Points Of Inflection

Example Question #2 : Determine Points Of Inflection

Example Question #1 : Determine Points Of Inflection

Example Question #1 : Determine Points Of Inflection

Example Question #116 : Derivatives

Example Question #117 : Derivatives

Example Question #118 : Derivatives

Example Question #1 : Determine Points Of Inflection

Example Question #2 : Determine Points Of Inflection

Example Question #1 : Determine Points Of Inflection

All Precalculus Resources

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