# Classifying Critical Points and Extreme Values for a Function

M7-1: extrema

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1 of 3 videos by Calculus V.

## Key Questions

• A critical point is when $f ' \left(x\right) \mathmr{and} f ' ' \left(x\right)$ is either equal to $0$, it does not exist (its a fraction and the bottom is equal to zero to create a vertical asymptote) or it is the original end points that were given.

So if you had the problem $y ' = \frac{8 - x}{22 + 11 x} ^ 2$, you would find the critical points by setting $8 - x = 0$ and ${\left(22 + 11 x\right)}^{2} = 0$. The first would be when the function is equal to 0 and the second is when the function does not exist because the bottom of a fraction cannot equal 0. This problem did not have endpoints or boundaries indicating where this graph starts and ends.

Critical points are used to find where the function is increasing, decreasing, maxims, points of inflection, and concave up and down. You can also use the information gathered from the critical points to draw a rough sketch of what your original function should look like.

• Here is how to find and classify a critical point of $f$.

Remember that $x = c$ is called a critical value of $f$ if $f ' \left(c\right) = 0$ or $f ' \left(c\right)$ is undefined.

$f ' \left(x\right) = 3 {x}^{2} = 0 R i g h t a r r o w x = 0$ is a critical number.

(Note: $f '$ is defined everywhere, $0$ is the only critical value.)

Observing that $f ' \left(x\right) = 3 {x}^{2} \ge 0$ for all $x$,

$f '$ does not change sign around the critical value $0$.

Hence, $f \left(0\right)$ is neither a local maximum nor a local minimum by First Derivative Test.

• Finding critical points of trig functions is not much different than the methods you use for finding critical points of any other types of functions. Critical points are points where the derivative = 0 or is undefined.

For example, consider
$f \left(x\right) = {\sin}^{2} \left(x\right)$ on the interval $\left[0 , 2 \pi\right\}$
We take the derivative using the chain rule:
$f ' \left(x\right) = 2 \sin x \cos x$
If we remember our trigonometric identities, we recognize this as the identity for $\sin \left(2 x\right)$ giving us:
$f ' \left(x\right) = \sin \left(2 x\right)$.

This function is never undefined because the domain of $y = \sin \left(\theta\right)$ is all real numbers. That leaves us finding points in $\left[0 , 2 \pi\right]$ where $f ' \left(x\right) = 0$

$0 = \sin \left(2 x\right)$
$2 x = 0 , \pi , 2 \pi , 4 \pi$
Dividing by two gives us
$x = 0 , \frac{\pi}{2} , \pi , 2 \pi$

Note that the first and last points are actually the endpoints of the domain. So, if we no wanted to find the intervals where the function is increasing/decreasing, we would test the value of the derivative in the three intervals determined by these critical points.

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