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Precalculus : Find the Critical Numbers of a Function

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Example Question #1 : Find The Critical Numbers Of A Function

What are the critical values of the function (x^3-2x^2+5) ?

Possible Answers:

None of the other answers

x=0

x=2

x=0, x=-2

$x=0;x=\frac{4}{3}$

Correct answer:

$x=0;x=\frac{4}{3}$

Explanation:

A number is critical if it makes the derivative of the expression equal 0.

Therefore, we need to take the derivative of the expression and set it to 0. We can use the power rule for each term of the expression.

$f^{'} = 3x^2 - 4x + 0 = 0$

Next, we need to factor the expression:

x(3x-4)= 0

We can now set each term equal to 0 to find the critical numbers:

x = 0

3x - 4 = 0

Therfore, our critical numbers are,

x=0

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

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Example Question #2 : Find The Critical Numbers Of A Function

Find the critical value(s) of the function.

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

Possible Answers:

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

$x=0,-\frac{1}{3}$

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

$x=\frac{1}{2},-\frac{1}{4}$

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

Correct answer:

$x=0,-\frac{1}{3}$

Explanation:

To find the critical values of a function, we must set the derivate equal to 0.

First, we find the derivative of the function to be

$\frac{dy}{dx}=3(6)x^2+2(3)x$

=18x^2+6x

We can then factor out a 6x and set the expression equal to 0

6x(3x+1)=0

From here, we can easily determine that

6x=0

x=0

and

3x+1=0

3x=-1

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

Therefore, the critical values of the function are at x=0 and $x=-\frac{1}{3}$ .

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Example Question #3 : Find The Critical Numbers Of A Function

Find the critical points of the following function:

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

Possible Answers:

x=-3,x=5

x=-5,x=3

x=-4,x=15

x=-4,x=5

x=-15,x=4

Correct answer:

x=-5,x=3

Explanation:

The critical points of a function are the points at which its slope is zero, so first we must take the derivative of the function so we have a function that describes its slope:

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

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

Now that we have the derivative, which tells us the slope of f(x) at any point x, we can set it equal to 0 and solve for x to find the points at which the slope of the function is 0, which are our critical points:

$x^2+2x-15=0\rightarrow (x+5)(x-3)=0\rightarrow x=-5,x=3$

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Example Question #4 : Find The Critical Numbers Of A Function

Find the critical number(s) of the function

f(x)=x^2 .

Possible Answers:

DNE

0.5

Correct answer:

Explanation:

To find the critical numbers, find the values for x where the first derivative is 0 or undefined.

For the function

f(x)=ax^n

The first derivative is

$f^{'}(x)=anx^{n-1}$

So for

f(x)=x^2

the first derivative is

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

Setting that equal to zero

2x=0

We find

x=0

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Example Question #5 : Find The Critical Numbers Of A Function

Find the critical number(s) of the function

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

Possible Answers:

0, 2

DNE

Correct answer:

0, 2

Explanation:

To find the critical numbers, find the values for x where the first derivative is 0 or undefined.

For the function

f(x)=ax^n .

The first derivative is

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

So for

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

the first derivative is

$f'(x)=3x^{3-1}-3(2)x^{2-1}=3x^2-6x=3x(x-2)$ .

Settig the first derivative equal to zero

f'(x)=3x(x-2) = 0

yields the values

x=0,2

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Example Question #1 : Find The Critical Numbers Of A Function

Find the critical numbers of the following function:

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

Possible Answers:

x=-6,x=2

x=5,x=7

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

x=-7,x=-5

x=-2,x=6

Correct answer:

x=-2,x=6

Explanation:

The critical numbers of a function are those at which its first derivative is equal to 0. These points tell where the slope of the function is 0, which lets us know where the minimums and maximums of the function are. First we find the derivative of the function, then we set it equal to 0 and solve for the critical numbers:

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

f'(x)=x^2-4x-12=(x-6)(x+2)=0

x=-2,x=6

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Example Question #7 : Find The Critical Numbers Of A Function

Find the critical numbers of the following function:

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

Possible Answers:

x=-3,x=-2

x=2,x=3

x=5,x=12

x=-12,x=5

$x=-\frac{2}{3},x=4$

Correct answer:

x=2,x=3

Explanation:

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

f'(x)=2x^2-10x+12=2(x^2-5x+6)=2(x-2)(x-3)=0

x=2,x=3

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Example Question #1 : Find The Critical Numbers Of A Function

Find the critical numbers of the following function:

$f(x)=x^3+\frac{3}{2}x^2-90x$

Possible Answers:

x=-5,x=6

$x=-\frac{3}{4},x=1$

x=-4,x=3

x=-6,x=5

x=-3,x=4

Correct answer:

x=-6,x=5

Explanation:

$f(x)=x^3+\frac{3}{2}x^2-90x$

f'(x)=3x^2+3x-90=3(x^2+x-30)=3(x+6)(x-5)=0

x=-6,x=5

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Example Question #1 : Find The Critical Numbers Of A Function

Find the critical value(s) of the function $f(x)=3x^{3}$ .

Possible Answers:

Answer not listed

x=0,-3

x=0,3

x=3,-3

x=0

Correct answer:

x=0

Explanation:

The critical values of a function f(x) are values for which the derivative f'(x)=0 . In this case:

$f(x)=3x^{3}$

$f'(x)=9x^{2}$

Setting f'(x)=0 :

$9x^{2}=0$

$x^{2}=0$

x=0

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Example Question #10 : Find The Critical Numbers Of A Function

Find the critical value(s) of the function $f(x)=x^{2}-2x+1$ .

Possible Answers:

x=1

x=1,-1

x=1,0

x=-1

Correct answer not listed

Correct answer:

x=1

Explanation:

The critical values of a function f(x) are values for which the derivative f'(x)=0 . In this case:

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

f'(x)=2x-2

Setting f'(x)=0 :

2x-2=0

2x=2

x=1

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Precalculus : Find the Critical Numbers of a Function

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