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Precalculus : Introductory Calculus

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Example Questions

Example Question #103 : Derivatives

Determine the critical numbers of the function

$\small f(x)=\frac{x^2}{2}-x$

Possible Answers:

$\small DNE$

$\small 1,0$

$\small 1$

$\small 2$

Correct answer:

$\small 1$

Explanation:

The critical numbers are the $\small c$ values for which either

$\small \small f'(c)=0$ or $\small \small f'(c)$ is undefined.

In order to find the first derivative we use the power rule which states

$\small \frac{\mathrm{d} }{\mathrm{d} x} x^n = nx^{n-1}$

Applying this rule term by term we get

$\small f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\frac{x^2}{2}-x]$

$\small =2\cdot\frac{x^{2-1}}{2}-1\cdot x^{1-1}$

$\small =x-1$

The first derivative is defined for all values of x. Setting the first derivative to zero yields

$\small f'(x)=x-1=0$

$\small x=1$

As such, the critical number is

$\small 1$

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Example Question #15 : Find The Critical Numbers Of A Function

Find the critical values of the following function:

f(x)=x^2-4

Possible Answers:

none

Correct answer:

Explanation:

To solve, simply find the first derivative and find when it is equal to 0. To find the first derivative, we must use the power rule as outlind below.

Power rule: $(x^n)'=nx^{n-1}$

Thus,

f(x)=x^2-4

$f'(x)=2x^{2-1}-4x^{1-1}=2x^1-4x^0=2x-4$

Now, we must set out function equal to 0 and solve for x. Thus,

2x=0

Dividing both sides by 2, we get:

x=0

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Example Question #111 : Derivatives

Find the critical values of the following function.

f(x)=x^2-6x+3

Possible Answers:

Correct answer:

Explanation:

To solve, simply differentiate using the power rule, as outlined below.

Power rule states,

$(x^n)'=nx^{n-1}$ .

Thus given,

f(x)=x^2-6x+3

our first derivative is:

$f'(x)=2x^{2-1}-6x^{1-1}=2x-6$

Then plug in 0 for f(x) to find when our function is equal to 0.

Thus,

0=2x-6

2x=6

x=3

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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=-3,x=4

x=-4,x=3

x=4,x=7

x=-2,x=5

x=-5,x=2

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 #1 : 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=-1,x=5

x=2,x=4

x=-3,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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Precalculus : Introductory Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #103 : Derivatives

Example Question #15 : Find The Critical Numbers Of A Function

Example Question #111 : Derivatives

Example Question #1 : Determine Points Of Inflection

Example Question #1 : 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

All Precalculus Resources

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