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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.

$\delta$ Neighborhood A $\delta$ neghborhood of $a$ is a set of real number $x$ such that $\vert x - a\vert < \delta$. In other words, $-\delta < x-a < \delta$.

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 $f^{\prime}(a) = 0$.

Understanding
Note that if a differentiable function $f$ takes local extrema at $x=a$. Then the slope of tangent line of $f$ at $x=a$ is 0.

On the other hand, consider $f(x) = x^3$. Since $f'(x) = 3x^2$, the slope of the tangent line of $f(x)$ is 0 at $x = 0$. But $f(x)$ does not take local extrema at $x = 0$.
A function may take local extrema without being differentiable. Consider $f(x) = \vert x\vert$.

Figure 2.10:
\includegraphics[width=4.2cm]{CALCFIG/Fig2-5-1.eps}

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

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

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

Understanding
Note that the 1st derivative represents the slope of the tangent line. Then the 2nd derivative represents how the slope of tangent line changes. If $f''(a) > 0$, then the slope of the tangent line is increasing in neighborhood of $a$ .

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 $f(x)$.

Extreme Point
1. Find a domian of a fucntion
2. Find a critical point .
3. Find a candidate for inflection point.
4. Draw a concavity table

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

$\displaystyle 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

$\displaystyle 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{displaymath}\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} \end{displaymath}

Check
How to find $f4. $. Write $f(x) = x^5 - 5x^4 + 1 = x^4(x-5) + 1$. Then substitute $x = 4$ to get $f4. = 4^4(4-5) + 1 = 256(-1) + 1 = -256+1 = -255$.

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 $\displaystyle{f(x) = \frac{x^3}{x^2 - 2}}$ and concavity of the graph of $f(x)$.

Since $f(x) = \frac{x^3}{x^2 - 2}$, ${\rm Dom}(f) = \{x : x \neq \pm \sqrt{2}\}$. SOLUTION By the quotient rule,
$\displaystyle y^{\prime}$ $\displaystyle =$ $\displaystyle \frac{(x^3)'(x^2 -2) - x^3(x^2 - 2)'}{(x^2 - 2)^2} = \frac{x^2 (x^2 - 6)}{(x^2 - 2)^2},$  
$\displaystyle y''$ $\displaystyle =$ $\displaystyle \frac{(x^2)'(x^2 -6) - x^2(x^2 - 6)'}{(x^2 -2)^4} = \frac{4x(x^2 + 6)}{(x^2 - 2)^3}$  

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

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


Subsections
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