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Precalculus : Pre-Calculus

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Example Questions

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:

$x=\pm1$

There are no inflection points.

$x=0,\pm1$

x=-1

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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Example Question #11 : Determine Points Of Inflection

Determine the point(s) of inflection of $y=\frac{1}{x+4}$ .

Possible Answers:

(0,0)

(-4,0)

$\left(0,\frac{1}{4}\right)$

No points of inflection exist.

(-5,-1) and (-3,1)

Correct answer:

No points of inflection exist.

Explanation:

The points of inflection exist where the second derivative is zero.

$y=\frac{1}{x+4}$

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

$y''=\frac{2}{(x+4)^3}$ which can never be . Therefore, there are no points of inflection.

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Example Question #12 : Determine Points Of Inflection

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

Possible Answers:

x=-3,3

No points of inflection exist.

x=0,3

x=3

x=-3

Correct answer:

No points of inflection exist.

Explanation:

The points of inflection will exist where the second derivative is zero.

$y=(x+3)^\frac{2}{3}$

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

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

$0=-\frac{2}{3}(x+3)^\frac{-4}{3}$

$0=-\frac{2}{3}\cdot\frac{1}{\sqrt[3]{x+3}^4}$ .

This will never be , so there are no points of inflection.

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Example Question #13 : Determine Points Of Inflection

List the interval(s) where the function y=x^3-2x^2+2 is concave up.

Possible Answers:

$\left(0,\frac{3}{2}\right)$

The graph is never concave up.

The graph is always concave up.

$\left(-\infty,\frac{2}{3}\right)$

$\left(\frac{2}{3},\infty\right)$

Correct answer:

$\left(\frac{2}{3},\infty\right)$

Explanation:

The graph is concave up where the second derivative is positive. Let us first find out if there are any points of inflection to narrow our search.

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

y'=3x^2-4x

y''=6x-4

0=6x-4

$x=\frac{2}{3}$ .

Now we can perform the second derivative concavity test on points on either side of $x=\frac{2}{3}$ . Let us try x=0,1 .

The second derivative at x=0 gives us y''=-4 which is less than zero, so the graph is concave down in this interval. The second derivative at x=1 gives us y''=2 , which is positive. Hence, the graph is concave up on the interval $\left(\frac{2}{3},\infty\right)$ .

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Example Question #13 : Determine Points Of Inflection

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

Possible Answers:

$x=\pm3$

x=0,3

There are no points of inflection.

$x=0,\sqrt{27}$

x=0

Correct answer:

There are no points of inflection.

Explanation:

The points of inflection will only exist where the second derivative is zero.

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

y'=2x

y''=2

Now $0\neq2$ therefore, there are no points of inflection.

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Example Question #11 : Determine Points Of Inflection

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

Possible Answers:

(2,3)

$\left(-1,\frac{1}{4}\right)$

(-2.28,-2.53),(0.92,3.92),(2.86,2.22)

$\left(-1,\frac{1}{4}\right),(2,3)$

There are no points of inflection.

Correct answer:

$\left(-1,\frac{1}{4}\right),(2,3)$

Explanation:

The points of inflection, if any exist, will be found where the second derivative is zero.

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

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

y''=x^2-x-2

0=x^2-x-2

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

x=-1,2 .

To find the y-coordinates, we simply plug the x-coordinates in to the original function.

$y=\frac{1}{12}(-1)^4-\frac{1}{6}(-1)^3-(-1)^2+2(-1)+3=\frac{1}{4}$ .

So $(-1,\frac{1}{4})$ is an inflection point.

Also, $y=\frac{1}{12}(2)^4-\frac{1}{6}(2)^3-(2)^2+2(2)+3=3$ , so (2,3) is another inflection point.

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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #11 : Determine Points Of Inflection

Example Question #12 : Determine Points Of Inflection

Example Question #13 : Determine Points Of Inflection

Example Question #13 : Determine Points Of Inflection

Example Question #11 : Determine Points Of Inflection

All Precalculus Resources

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