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Precalculus : Derivatives

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Example Questions

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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Example Question #791 : Pre Calculus

List the intervals and determine where the graph of $y=\frac{1}{3}x^4+\frac{1}{3}x^3-6x^2+x+5$ is concave up and concave down.

Possible Answers:

The graph is concave up on $\left(-2,\frac{3}{2}\right)$ .

The graph is concave down on $(-\infty,-2)$ and $\left(\frac{3}{2},\infty\right)$ .

The graph is concave up on $(-\infty,0)$ .

The graph is concave down on $(0,\infty)$ .

The graph is always concave up.

The graph is concave up on $(-\infty,-2)$ and $\left(\frac{3}{2},\infty\right)$ .

The graph is concave down on $\left(-2,\frac{3}{2}\right)$ .

The graph is always concave down.

Correct answer:

The graph is concave up on $(-\infty,-2)$ and $\left(\frac{3}{2},\infty\right)$ .

The graph is concave down on $\left(-2,\frac{3}{2}\right)$ .

Explanation:

We need to set the second derivative equal to zero to determine where the inflection points are.

$y=\frac{1}{3}x^4+\frac{1}{3}x^3-6x^2+x+5$

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

y''=4x^2+2x-12

0=4x^2+2x-12

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

0=2x^2+x-6

0=(x+2)(2x-3)

$x=-2,\frac{3}{2}$ are the x-coordinates of our inflection points. Thus the intervals of concavity are $(-\infty,-2)$ , $\left(-2,\frac{3}{2}\right)$ , and $\left(\frac{3}{2},\infty\right)$ . We can use x=-3,0,2 as our test points.

y''(-3)=18>0 , so the graph is concave up on $(-\infty,-2)$ .

y''(0)=-12< 0 , so the graph is concave down on $\left(-2,\frac{3}{2}\right)$ .

y''(2)=8>0 , so the graph is concave up on $\left(\frac{3}{2},\infty\right)$ .

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Example Question #11 : Determine Points Of Inflection

Determine the points of inflection, if any, of the following function:

f(x)=x^3-5x^2+2x+11

Possible Answers:

$x=\frac{5}{3}$

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

x=4

x=-7

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

Correct answer:

$x=\frac{5}{3}$

Explanation:

The points of inflection of a function are those at which its concavity changes. The concavity of a function is described by its second derivative, and when the second derivative is 0 a point of inflection occurs. We find the second derivative of the function and then set it equal to 0 to solve for the inflection points:

f(x)=x^3-5x^2+2x+11

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

f''(x)=6x-10

$6x-10=0\rightarrow 6x=10\rightarrow x=\frac{5}{3}$

So the function has only one point of inflection at x=5/3.

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Example Question #793 : Pre Calculus

Which of the following is an -coordinate of an inflection point of the graph of the following function?

$f(x)=\frac{1}{12}x^4-2x^3+16x^2-19x+36$

Possible Answers:

$-\frac{2}{3}$

Correct answer:

Explanation:

The inflection points of a function are the points where the concavity changes, either from opening upwards to opening downwards or vice versa. The inflection points occur at the x-values where the second derivative is either zero or undefined. That means we need to find our second derivative.

We start by using the Power Rule to find the first derivative.

$f'(x)=\frac{1}{3}x^3-6x^2+32x-19$

Taking the derivative once more gives the second derivative.

f''(x)=x^2-12x+32

We then set this derivative equal to zero and solve.

x^2-12x+32=0

This factors nicely.

(x-8)(x-4)=0

Therefore our second derivative is zero when $x=4 \: or \: 8$

8 is the only one of these two amongst our choices and is therefore our answer.

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Example Question #791 : Pre Calculus

Determine the points of inflection for the following function.

f(x)=3x^2+4x-2

Possible Answers:

None

Correct answer:

None

Explanation:

To find inflection points, take the second derivative and set it equal to .

f'(x)=6x+4

f''(x)=6

Since is never , this function does not have any inflection points. Thus, the answer is none.

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Precalculus : Derivatives

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

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

Example Question #791 : Pre Calculus

Example Question #11 : Determine Points Of Inflection

Example Question #793 : Pre Calculus

Example Question #791 : Pre Calculus

All Precalculus Resources

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