Calculus/Extreme Value Theorem

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 Test
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.

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

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: $$ x = 0 \ $$.