# Calculus/Extreme Value Theorem

 ← Euler's Method Calculus Rolle's Theorem → Extreme Value Theorem
Extreme Value Theorem
If f is a continuous function and closed on the interval [$a,b$], then f has both a minimum and a maximum.

This introduces us to the aspect of global extrema and local extrema. (Also known as absolute extrema or relative extrema respectively.)

How is this so? Let us use an example.

$f(x) = x^2$ and is closed on the interval [-1,2]. Find all extrema.

$\frac{dy}{dx} = 2x$

A critical point (a point where the derivative is zero) exists at (0,0). Just for practice, let us use the second derivative test to evaluate whether or not it is a minimum or maximum. (You should know it is a minimum from looking at the graph.)

$\frac{d^2y}{dx^2} = 2$

$f''(c) > 0$, thus it must be a minimum.

As mentioned before, one can find global extrema on a closed interval. How? Evaluate the y coordinate at the endpoints of the interval and compare it to the y coordinates of the critical point. When you are finding extrema on a closed interval it is called a local extremum and when it's for the whole graph it's called a global extremum.

1: Critical Point: (0,0) This is the lowest value in the interval. Therefore, it is a local minimum which also happens to be the global minimum.

2: Left Endpoint (-1, 1) This point is not a critical point nor is it the highest/lowest value, therefore it qualifies as nothing.

3: Right Endpoint (2, 4) This is the highest value in the interval, and thus it is a local maximum.

This example was to show you the extreme value theorem. The quintessential point is this: on a closed interval, the function will have both minima and maxima. However, if that interval was an open interval of all real numbers, (0,0) would have been a local minimum. On a closed interval, always remember to evaluate endpoints to obtain global extrema.

## First Derivative TestEdit

Recall that the first derivative of a function describes the slope of the graph of the function at every point along the graph for which the function is defined and differentiable.

Increasing/Decreasing:

• If $f'(x) < 0 \$, then $f(x) \$ is decreasing.
• If $f'(x) > 0 \$, then $f(x) \$ is increasing.

Local Extrema:

• If $\frac{dy}{dx}|_{x=c} = f'(c) = 0$ and $f'(x) \$ changes signs at $x = c \$, then there exists a local extremum at $x = c \$.
• If $f'(x) < 0 \$ for $x < c \$ and $f'(x) > 0 \$ for $x > c \$, then $f(c) \$ is a local minimum.
• If $f'(x) > 0 \$ for $x < c \$ and $f'(x) < 0 \$ $x > c \$, then $f(c) \$ is a local maximum.

Example 1:

Let $f(x) = 3x^2 + 4x - 5 \$. Find all local extrema.

• Find $\frac{dy}{dx}$
$f(x) = 3x^2 + 4x - 5 \$
$f'(x) = 6x + 4 \$
• Set $\frac{dy}{dx} = 0$ to find local extrema.
$6x + 4 = 0 \$
$6x = -4 \$
$x = - \frac{2}{3}$
• Determine whether there is a local minimum or maximum at $x = - \frac{2}{3}$.
Choose an x value smaller than $- \frac{2}{3}$:
$f'(-1) = 6(-1) + 4 = -2 < 0 \$
Choose an x value larger than $- \frac{2}{3}$:
$f'(1) = 6(1) + 4 = 10 > 0 \$

Therefore, there is a local minimum at $x = - \frac{2}{3}$ because $f'(- \frac{2}{3}) = 0$ and $f'(x) \$ changes signs at $x = - \frac{2}{3}$.

Answer: local minimum: $x = - \frac{2}{3}$.


## Second Derivative TestEdit

Recall that the second derivative of a function describes the concavity of the graph of that function.

• If $\frac{d^2y}{dx^2}|_{x=c} = f''(c) = 0$ and $f''(c) \$ changes signs at $x = c \$, then there is a point of inflection (change in concavity) at $x = c \$.
• If $f''(x) < 0 \$, then the graph of $f(x) \$ is concave down.
• If $f''(x) > 0 \$, then the graph of $f(x) \$ is concave up.

Example 2:

Let $f(x) = x^3 + 2x + 7 \$.  Find any points of inflection on the graph of $f(x) \$.

• Find $\frac{d^2y}{dx^2}$.
$f(x) = x^3 + 2x + 7 \$
$f'(x) = 3x^2 + 2 \$
$f''(x) = 6x \$
• Set $\frac{d^2y}{dx^2} = 0$.
$6x = 0 \$
$x = 0 \$
• Determine whether $f''(x) \$ changes signs at $x = 0 \$.
Choose an x value that is smaller than 0:
$f''(-1) = 6(-1) = -6 < 0 \$
Choose an x value that is larger than 0:
$f''(1) = 6(1) = 6 > 0 \$

Therefore, there exists a point of inflection at $x = 0 \$ because $f''(0) = 0 \$ and $f''(x) \$ changes signs at $x = 0 \$.

Answer: point of inflection: $x = 0 \$.