Extreme Value of Functions

Extreme Value

For all $x$ in the neighborhood of $a$, $f(x) < f(a)$. Then we say $f(x)$ takes local maximum at $x = a$. if $f(x) > f(a)$, then we say $f(x)$ takes local minimum at $x = a$. local maximum and local minimum together called local extrema.

First Derivative Test

Theorem 2..18   Suppose that $f(x)$ is differentiable at $x = a$. If $f(x)$ takes local extrema at $x = a$, then $f^{\prime}(a) = 0$.

NOTE If $f^{\prime}(a) > 0$, then $f(x)$ is increasing at $x = a$. If $f^{\prime}(a) < 0$ , then $f(x)$ is decreasing at $x = a$ . In these cases, $f(x)$ does not take local extrema. Thus we must have $f^{\prime}(a) = 0$.

Criterion for Local Extrema

Theorem 2..19   Suppose that $f(x)$ is continuous on a neighborhood of $x = a$. If $h > 0$ is small enough, then

1.	If $f^{\prime}(x) > 0$ on $(a-h,a)$ and $f^{\prime}(x) < 0$ on $(a,a+h)$,
	then $f(x)$ takes local maximum at $x = a$.
2.	If $f^{\prime}(x) < 0$ on $(a-h,a)$ and $f^{\prime}(x) > 0$ on $(a,a+h)$,
	then $f(x)$ takes local minimum at $x = a$.
3.	If $f^{\prime}(x)$ does not change the sign on $(a-\delta, a + \delta)$,
	then $f(a)$ is not local extrema.

NOTE 1. $f(x)$ is strictly increasing function on $[a-h,a]$ and strictly decreasin on $[a,a+h]$. Thus $f(x)$ takes the local maximum at $x = a$.

Inflection Point

an inflection point, point of inflection, flex, or inflection (inflexion) is a point on a curve at which the curvature or concavity changes sign from plus to minus or from minus to plus..

2nd Derivative Test

Theorem 2..20   Suppose $f(x)$ is twice differentialbe on a interval containing $x = a$ and satisfies $f^{\prime}(a) = 0$.

If $${1. \ f^{\prime\prime}(a) > 0}$$, then the graph of $f$ is concave up, and $f(a)$ is local minimum

If $${2. \ f^{\prime\prime}(a) < 0}$$, then the graph of $f$ is concave down, and $f(a)$ is local maximum

If $${3. \ f^{\prime\prime}(a) = 0}$$ and concavity changes, then $(a,f(a))$ is an inflection point.

NOTE Apply the above theorem to a function $f^{\prime}(x)$, Then $f^{\prime}(x)$ is increasing at $x = a$. Since $f^{\prime}(a) = 0$, $f^{\prime}(x)$ takes negative on $(a - \delta,a)$ on positive on $(a, a+\delta)$ . Thus the graph of a function is concave up at $x = a$ and takes a local minimum at $x = a$.

Example 2..18   Find a extreme point of $f(x) = x^5 - 5x^4 + 1$ and concavity of the graph of $f(x)$.

SOLUTION Since $f(x)$ is differentiable on $(-\infty,\infty)$, if $f(x)$ attains etremumat some point, then at the point $f^{\prime}(x) = 0$. Thus, we find $x$ so that $f^{\prime}(x) = 0$. Since

$$f^{\prime}(x) = 5x^4 - 20x^3 = 5x^3(x - 4) = 0,$$

$x = 0,4$ are the candidate for critical point. Next to check concavity of the graph of $f$, we find $f^{\prime\prime}(x)$. Since

$$f^{\prime\prime}(x) = 20x^3 -60 x^2 = 20x^2 (x - 3)$$

$x = 0,3$ are the candidate for inflection point. Now we create a concavity table.

$$\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...\rm IP} & {\rm up} & {\rm up} & {\rm up} \\ \hline \end{array}$$

By the 1st derivative test, $f(0) = 1$ is a local maximum. $f4. = -255$ is a local minimum. By the 2nd derivative test, $(3,f3. )$ is an inflection point. The graph of function is concave down on the left-hand side of the inflection point and concave up on the right-hand side of the inflection point $\ \blacksquare$

Exercise 2..18   Find a extreme point of the function $${f(x) = \frac{x^3}{x^2 - 2}}$$ and concavity of the graph of $f(x)$.

SOLUTION By the quotient rule,

$$y^{\prime} = \frac{(x^3)'(x^2 -2) - x^3(x^2 - 2)'}{(x^2 - 2)^2} = \frac{x^2 (x^2 - 6)}{(x^2 - 2)^2},$$

$$y'' = \frac{(x^2)'(x^2 -6) - x^2(x^2 - 6)'}{(x^2 -2)^4} = \frac{4x(x^2 + 6)}{(x^2 - 2)^3}$$

Then $x = 0, \pm \sqrt{6}$ are candidates for a critical point. Now write a concavity table.

$$\begin{array}{c\vert ccccccccccccc} x & & -\sqrt{6} & & - \sq... ...3\sqrt{6}}{2} & \nearrow \end{array}\ensuremath{\ \blacksquare}$$