Find the Critical Numbers of a Function - Pre-Calculus

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Question

What are the critical values of the function ?

Answer

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,

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

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Question

Find the critical value(s) of the function.

Answer

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

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

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

From here, we can easily determine that

and

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

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

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Question

Find the critical points of the following function:

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

Answer

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:

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

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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Question

Find the critical number(s) of the function

.

Answer

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

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

Setting that equal to zero

We find

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Question

Find the critical number(s) of the function

.

Answer

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

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Question

Find the critical numbers of the following function:

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

Answer

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:

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

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Question

Find the critical numbers of the following function:

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

Answer

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:

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

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Question

Find the critical numbers of the following function:

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

Answer

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:

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

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Question

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

Answer

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

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

Setting :

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Question

Find the critical numbers of the following function:

Answer

The critical numbers of a function are the points at which its slope is zero, which tells where the function has a minimum or maximum. The slope of a function is described by its derivative, so we'll take the derivative of the function and set it equal to 0 to solve for the x values of the critical numbers:

It appears the derivative cannot be factored to solve for x, so we'll have to use the quadratic formula to find the critical numbers:

$\frac{10\pm \sqrt{(-10)^2-4(3)(4)}}{2(3)}=\frac{5}{3}\pm\frac{1}{3}\sqrt{13}=\frac{1}{3}(5\pm \sqrt{13})$

So the critical numbers occur at the following two values of x:

$x=\frac{1}{3}(5-\sqrt{13}),x=\frac{1}{3}(5+\sqrt{13})$

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Question

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

Answer

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

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

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

Setting :

$9x^{2}=0$

$x^{2}=0$

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Question

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

Answer

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 :

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

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

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Question

Given the following function, find the critical numbers:

Answer

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

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Question

Determine the critical numbers of the function

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

Answer

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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Question

Find the critical values of the following function.

Answer

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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Question

Find the critical values of the following function:

Answer

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,

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

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

Dividing both sides by 2, we get:

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