3.1 Critical Numbers

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In this section we learn to find the critical numbers of a function.

Critical Numbers

In this section we will find the critical numbers of a given function. Critical numbers come in two main types and the idea of the definition is to consider the possibilities at a local extreme. Here is the definition of critical number.

A critical number for the function $f(x)$ is a number $x_0$ in the domain of the function $f(x)$ such that either $f'(x_0) = 0 \text { (type 1)}$ $\text {or}$ $f'(x_0) \text { is undefined (type 2)}.$

In other words, at a critical number $x_0$ we have $f(x_0)$ is defined and either $f'(x_0) = 0$ or $f(x)$ is not differentiable at $x_0$ .

For an example of the first type consider the function $f(x) = x^2$ . Since $f'(x) = 2x$ which equals 0 when $x = 0$ , $f(x)$ has a critical number at $x = 0$ .

For an example of the second type consider the function $f(x) = |x|$ . This function has a corner point at $x = 0$ so $f(x)$ is not differentiable there. Hence $x=0$ is a critical number for $f(x)$ .

Examples of Finding Critical Numbers

CN 1.

Find the critical numbers of the function $f(x) = \dfrac {x^3}{3} - \dfrac {x^2}{2} - 6x - 3.$ Solution: We need to compute $f'(x)$ . We have $f'(x) = x^2 - x - 6.$ Noting that $f'(x)$ is defined for all values of $x$ , there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $x^2 - x - 6 =0$ $(x+2)(x-3) =0$ $x = -2 \mbox { or } x = 3.$

Hence $f(x) = \frac {x^3}{3} - \frac {x^2}{2} - 6x - 3$ has two critical numbers, $-2$ and $3$ , and they are both type 1.

Find the critical numbers of the function $f(x) = x^2 - 8x + 11$ .

Compute $f'(x)$

The derivative is $f'(x) = 2x -8$

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = x^2 - 8x + 11$ are $\answer [given]{4}$

Find the critical numbers of the function $f(x) = 2x^3 - 3x^2 -36x + 7$

Compute $f'(x)$

The derivative is $f'(x) = 6x^2 - 6x - 36$

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = 2x^3 - 3x^2 -36x + 7$ are $\answer [given]{-2}$ and $\answer {3}$ (list answers in ascending order).

Find the critical numbers of the function $f(x) = x^4 - 4x^3 - 5$

Compute $f'(x)$

The derivative is $f'(x) = 4x^3 - 12x^2$

Solve $f'(x) = 0$ for $x$

Factor out $4x^2$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = x^4 - 4x^3 - 5$ are $\answer [given]{0}$ and $\answer {3}$ (list answers in ascending order).

CN 2.

Find the critical numbers of the function $f(x) = x^2e^{3x}.$ We need to compute $f'(x)$ using the product and chain rules. We have $f'(x) = (3x^2 +2x)e^{3x} \mbox { (verify)}.$ Noting that $f'(x)$ is defined for all values of $x$ , there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $(3x^2 +2x)e^{3x} =0$ $x(3x+2)e^{3x} =0$ $x = 0 \mbox { or } x = -2/3.$ Note that the equation $e^{3x} = 0$ has no solutions since an exponential function is always positive.

Hence $f(x) = x^2e^{3x}$ has two critical numbers, $-2/3$ and $0$ , and they are both type 1.

Find the critical numbers of the function $f(x) = xe^{2x}$

Compute $f'(x)$ using the Product Rule

The derivative can be written as $f'(x) = (2x + 1)e^{2x}$

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = xe^{2x}$ are $\answer [given]{-1/2}$

Find the critical numbers of the function $f(x) = 5xe^{-x/5}$

Compute $f'(x)$ using the Product Rule

The derivative can be written as $f'(x) = (-x + 5)e^{-x/5}$

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = 5xe^{-x/5}$ are $\answer [given]{5}$

CN 3.

Find the critical numbers of the function $f(x) = \dfrac {x}{x^2 +1}.$ We need to compute $f'(x)$ using the quotient rule. We have $f'(x) = \frac {1-x^2}{(x^2+1)^2} \mbox { (verify)}.$ Noting that $f'(x)$ is defined for all values of $x$ (since the denominator is never equal to 0), there are no type 2 critical numbers. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $\frac {1-x^2}{(x^2+1)^2} =0$ $1-x^2 =0$ $x = \pm 1.$

Hence $f(x) = \dfrac {x}{x^2 +1}$ has two critical numbers, $-1$ and $1$ , and they are both type 1.

Find the critical numbers of the function $f(x) = \dfrac {1}{1 + x^2}$

Compute $f'(x)$ using the Quotient Rule

The derivative can be written as $f'(x) = -\dfrac {2x}{(1 + x^2)^2}$

Solve $f'(x) = 0$ for $x$

Set the numerator to zero

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = \dfrac {1}{1 + x^2}$ are $\answer [given]{0}$

Find the critical numbers of the function $f(x) = \dfrac {3x}{x^2 +4}$

Compute $f'(x)$ using the Quotient Rule

The derivative can be written as $f'(x) = \dfrac {-3x^2 + 12}{(x^2 + 4)^2}$

Solve $f'(x) = 0$ for $x$

Set the numerator to zero

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = \dfrac {3x}{x^2 + 4}$ are $\answer [given]{-2}$ and $\answer {2}$ (input answers in ascending order).

CN 4.

Find the critical numbers of the function $f(x) = \sqrt [3] x.$ We need to compute $f'(x)$ . We have $f'(x) = \frac {1}{3\sqrt [3]{x^2}} \text { (verify)}.$ In this case, $f'(0)$ is undefined (division by zero). Hence $x=0$ is a critical number if $f(0)$ is defined. We can easily check this: $f(0) = \sqrt [3] 0 = 0$ , so it is defined and now we can conclude that $x=0$ is a type 2 critical number. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $\frac {1}{3\sqrt [3]{x^2}} =0$ $1 =0.$ So there are no solutions. The function has no type 1 critical numbers.

Hence $f(x) = \sqrt [3] x$ has only one critical number, 0, and it type 2, where the function is not differentiable. Geometrically, the function $f(x) = \sqrt [3] x$ has a vertical tangent line at the critical number $x = 0$ .

Find the critical numbers of the function $f(x) = \sqrt [5] x$

Compute $f'(x)$ using the Power Rule

The derivative can be written as $f'(x) = \dfrac {1}{5x^{4/5}}$

Solve $f'(x) = 0$ for $x$

No solutions!

Are there any points where $f'(x)$ is undefined? (Denominator equals zero!) If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = \sqrt [5] x$ are $\answer [given]{0}$

Find the critical numbers of the function $f(x) = x\sqrt [3]x$

Rewrite $f(x)$ as a power function by adding exponents

The derivative can be written as $f'(x) = \frac 43 x^{1/3}$

Solve $f'(x) = 0$ for $x$

Are there any points where $f'(x)$ is undefined? If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = x\sqrt [3]x$ are $\answer [given]{0}$

CN 5.

Find the critical numbers of the function $f(x) = \dfrac {1}{x}.$ We need to compute $f'(x)$ . We have $f'(x) = -\frac {1}{ x^2} \mbox { (verify)}.$ In this case, $f'(x)$ is undefined at $x = 0$ (division by zero). Hence, if $f(0)$ is defined then $x=0$ would be a type 2 critical number. However, we can easily see that $f(0) = \dfrac {1}{0}$ , is undefined, so that $x=0$ is a not in the domain of $f(x)$ and hence it is not a critical number. To find the type 1 critical numbers, we solve the equation $f'(x) = 0.$ Geometrically, these are the points where the graph of $f(x)$ has horizontal tangent lines. We get $-\frac {1}{ x^2} =0$ $1 =0.$ So there are no solutions. The function has no type 1 critical numbers.

Hence $f(x) = \dfrac {1}{x}$ has no critical numbers.

Find the critical numbers of the function $f(x) = \dfrac {3}{x^2}$

Compute $f'(x)$ using the Power Rule

The derivative can be written as $f'(x) = -\dfrac {6}{x^3}$

Solve $f'(x) = 0$ for $x$

If there are no solutions, write none

Are there any points where $f'(x)$ is undefined? (Denominator equals zero!) If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = 3/x^2$ are $\answer [given]{none}$ If there are none, type “none”.

Find the critical numbers of the function $f(x) = \tan (x)$

Recall $f'(x) = \sec ^2(x)$

Solve $f'(x) = 0$ for $x$

No solutions! (Note that $\sec (x) = \frac {1}{\cos (x)}$ )

Are there any points where $f'(x)$ is undefined? (Denominator equals zero!) If so, is $f(x)$ defined at these points?

The critical numbers of $f(x) = \tan (x)$ are $\answer [given]{none}$

If there are none, type “none”.

Here is a detailed, lecture style video on critical numbers: