Find the Critical Numbers of a Function

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Pre-Calculus › Find the Critical Numbers of a Function

Questions 1 - 10

1

Given the following function, find the critical numbers:

$\frac{1}{3}$

$-\frac{1}{3}$

$-\frac{2}{3}$

Explanation

Critical numbers are where the slope of the function is equal to zero or undefined.

Find the derivative and set the derivative function to zero.

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

There is only one critical value at $x=\frac{1}{3}$ .

2

Find the critical number(s) of the function

.

DNE

Explanation

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

For the function

.

The first derivative is

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

So for

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

yields the values

3

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

Correct answer not listed

Explanation

The critical values of a function are values for which the derivative . In this case:

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

Setting :

4

Find the critical numbers of the following function:

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

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

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:

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

5

Find the critical numbers of the following function:

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

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

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:

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

6

What are the critical values of the function ?

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

None of the other answers

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:

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

Therfore, our critical numbers are,

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

7

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

Answer not listed

Explanation

The critical values of a function are values for which the derivative . In this case:

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

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

Setting :

$9x^{2}=0$

$x^{2}=0$

8

Find the critical numbers of the following function:

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

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

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:

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

9

Find the critical value(s) of a function $f(x)=(x+1)^{3}$ .

Explanation

The critical values of a function are values for which the derivative . In this case:

$f(x)=(x+1)^{3}$

$f'(x)=3(x+1)^{2}$

Setting :

$3(x+1)^{2}=0$

$(x+1)^{2}=0$

10

Find the critical values of the following function.

Explanation

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

Power rule states,

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

Thus given,

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,

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