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 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.)

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

$f''(c)>0$, 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.

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 Test

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

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

$f(x)=x^{3}+2x+7\$

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

$f''(x)=6x\$

• Set ${\frac {d^{2}y}{dx^{2}}}=0$.

$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\$.