Calculus/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 "absolute" extrema and "relative" extrema.

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

$\frac{d^{2}y}{d{x}^2}$ = 2. $f^{″}(c)>0$, thus it must be a minimum.

As I mentioned before, one can find absolute min/max's 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 min/max's on a closed interval it is called relative min/max and when it's for the whole graph it's called Absolute min/max.

1: Critical Point: (0,0) This is the lowest value in the interval. Therefore, it is an "Relative Minimum" which also happens to be the absolute 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 an "Relative 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 minimums and maximums. However, if that interval was an open interval of all real numbers, (0,0) would have been a relative minimum. On a closed interval, always remember to evaluate endpoints to obtain Absolute 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.

Increasing/Decreasing:

• If $f^{\prime }(x)<0$, then $f(x)$ is decreasing.

• If $f^{\prime }(x)>0$, then $f(x)$ is increasing.

Relative Extrema:

• If $\frac{dy}{dx}{|}_{x=c}=f^{\prime }(c)=0$ and $f^{\prime }(x)$ changes signs at $x=c$, then there exists a relative extremum at $x=c$.

• If $f^{\prime }(x)<0$ for $x<c$ and $f^{\prime }(x)>0$ for $x>c$, then $f(c)$ is a relative minimum.

• If $f^{\prime }(x)>0$ for $x<c$ and $f^{\prime }(x)<0$ $x>c$, then $f(c)$ is a relative maximum.

Example 1:

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

• Find $\frac{dy}{dx}$

$f(x)=3x^2+4x-5$

$f^{\prime }(x)=6x+4$

• Set $\frac{dy}{dx}=0$ to find relative extrema.

$6x+4=0$

$6x=-4$

$x=-\frac{2}{3}$

• Determine whether there is a relative minimum or maximum at $x=-\frac{2}{3}$.

Choose a x value smaller than $-\frac{2}{3}$:

$f^{\prime }(-1)=6(-1)+4=-2<0$

Choose a x value larger than $-\frac{2}{3}$:

$f^{\prime }(1)=6(1)+4=10>0$

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

Answer: relative 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.

• If $\frac{d^{2}y}{d{x}^2}{|}_{x=c}=f^{″}(c)=0$ 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 $\frac{d^{2}y}{d{x}^2}$.

$f(x)=x^3+2x+7$

$f^{\prime }(x)=3x^2+2$

$f^{″}(x)=6x$

• Set $\frac{d^{2}y}{d{x}^2}=0$.

$6x=0$

$x=0$

• Determine whether $f^{″}(x)$ changes signs at $x=0$.

Choose a x value that is smaller than 0:

$f^{″}(-1)=6(-1)=-6<0$

Choose a 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: $x=0$.