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Using derivative to find extreme values

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We know that it might be possible to find maximum and minimum values of functions, if the functions are continuous.

Now practically, how can we use the derivative, $f^{\prime}$, to find global max/min?

Minimum and Maximum Points on a Graph.

Recall, these special points have horizontal tangents or they have no tangents at all.

So, we need to find all $x$'s such that

  • $f^{\prime}(x) = 0$, or
  • $f^{\prime}(x)$ does not exist, or
  • $x$ is $a$ or $b$ (if the domain is $[a,b]$ etc).

But what exactly does this give us? The derivative $f^{\prime}(c)$ is defined as a limit with $x$ approaching $c$. This is why it reflects only the behavior of $f$ close to $c$:

CandC.png

The idea of getting only the values of $f(x)$ for $x$ close to $c$ is presented more precisely in

Definition 
A function $f$ has a local minimum point at $x = c$ if
  • $f(a) \leq f(x)$ for all $x$ ... (no not all!)

in some interval $(a, b)$ that contains $c$. ("some" = "there is")

Example: Local Max and Min

Simple behavior in these intervals.
Definition 
A function $f$ has a local maximum point at $x = c$ if
  • $f(a) \geq f(x)$ for all $x$ in some interval $(a, b)$ that contains $c$.

We have found above: local max, min, max, min, etc. Around each of these points we see:

  • $\nearrow \searrow$ or $\nearrow \searrow$.

They correspond to max or min respectively.

Now, how do we find these points?

Claim: $f^{\prime}(c) = 0.$

Claim.png

Why? If not: $f^{\prime}(c) > 0$ or $f^{\prime}(c) < 0$. And there is no max or min!

Not max/min

"Formal" proof next.

Not max/min

Recall, $f^{\prime}(c)$ is the limit of slopes of the secant lines.

Consider secant lines on

  • A: the left of $c$ (green): $\text{slope} \geq 0$.
  • B: The right of $c$ (orange): $\text{slope} \leq 0$.

Recall: $$g(x) \geq 0 \Rightarrow \lim\limits_{x \to c} g(x)\geq 0.$$

We use this fact to conclude:

  • A. implies that $f^{\prime}(c) \geq 0$.
  • B. implies that $f^{\prime}(c) \leq 0.$

Simultaneously!

Therefore, $$f^{\prime}( c ) = 0.$$

To complete proof, we need to prove algebraically that $\text{slopes of secants } \geq 0, \leq 0$.

Take any (secant) line through $(x, f(x))$ and $(c,f(c)), x < c$. Then $$ f(x) \leq f(c)$$ if $x$ is close enough to $c$. Why? Because $c$ is a local max (review the definition). Now $$ \text{Slope } = \frac{\overbrace{f(c) - f(x)}^{\text{positive}}}{\underbrace{c - x}_{\text{positive}}}.$$ Observe that the numerator and denominator are positive, so the slope is positive.

Now take any (secant) line through $(x, f(x))$ and $(c,f(c)), x > c$. Then $$ f(x) \leq f(c) $$ if $x$ is close enough to $c$. Vecause $c$ is a local max. Then $$ \text{Slope } = \frac{\overbrace{f(c) - f(x)}^{\text{positive}}}{\underbrace{c - x}_{\text{negative}}}$$ Observe that the numerator and denominator are positive, so the slope is negative. $\blacksquare$

Let's formalize this result.

Fermat’s Theorem. Suppose $c$ is a local extreme point of $y = f(x)$. Then $$\left.\begin{aligned} f^{\prime}(c) & = 0 \quad \text{ or } \\ f^{\prime}(c) & \text{ does not exist.} \end{aligned} \right\} $$ In this case $c$ is is called a critical point.

Plan: To find global max/min, find all critical points.

Q: What about vice versa? Is the "converse" below of the theorem true?

$f^{\prime}(c) = 0$ then $c$ is a local extreme point of $f$.

A: Not true!

Example. Consider $f(x) = x^{3}$ at $x = 0$

No Extreme Points in Graph

$$\begin{aligned} f^{\prime}(x) &= 3x^{2} \\ f^{\prime}(0) & = 0, \end{aligned}$$ but this is not an extreme point.

Example. A more extreme example...

Local Max and Min

Verify.

Example: Global Extreme Points.

Find global extreme points of $y = f(x) = x^{3} - x$ on $[-2, 5]$.

Step 1: Find all critical points.

How? Use Fermat's theorem.

Compute: $$ f^{\prime}(x) = 3x^{2} - 1 $$ Set it to 0, and solve for $x$. $$\begin{aligned} 3x^{2} - 1 &= 0 \\ x^{2} &= \frac{1}{3} \\ x & = \pm \frac{1}{\sqrt{3}} \end{aligned}$$ There are 2 critical points.

Step 2: Compare the values of $f$ at these points and the end points of the interval, find the smallest and the largest:

$ $ $x$ $y = x^{3} - x = x(x^{2} - 1)$
Critical Points $\frac{1}{\sqrt{3}}$ $\frac{1}{\sqrt{3}}\left(\frac{1}{3} - 1 \right) = -\frac{2}{3\sqrt{3}}$
Critical Points $\frac{1}{\sqrt{3}}$ $-\frac{1}{\sqrt{3}}\left(-\frac{1}{3} - 1 \right) = \frac{2}{3\sqrt{3}}$
End Points $-2$ $8 + 2 = -5 \gets \text{ global min }$
End Points $5$ $12.5 - 5 = 120 \gets \text{ global max }$

The critical points are visible on the graph:

Step2.png

Recall: $$\lim_{x \to \pm\infty}(x^{3} - x) = \pm\infty.$$



Example: No Local Max/Min

No Local Max/Min Point

Plot with no local max/min showing critical points at 2, 4. Graph is continuous on $[1,5]$.

Exercise. Analyze the function (algebraically): $$f(x) = 2x^{3} - 3x^{2} - 12x + 1 $$ on $[-2,3]$.

Compute: $$f^{\prime}(x) = 6x^{2} - 6x - 12 $$ defined for all $x$. Solve $$\begin{aligned} 6x^{2} - 6x - 12 &= 0 \\ x^{2} -x - 2 &= 0 \\ \text{QF} \qquad x &= \frac{1 \pm \sqrt{1 - (-2)4}}{2} \\ &= \frac{1 \pm 3}{2} = 2,1 \end{aligned}$$

Compare:

$ $ $x$ $y$
Critical Points 2 $2\cdot 2^{3} - 3 \cdot 2^{2} - 12 \cdot 2 + 1 = 16 - 12 - 24 + 1 = -19$
Critical Points -1 $2\cdot (-1)^{3} - 3 \cdot (-1)^{2} - 12 \cdot (-1) + 1 = -2 - 3 +12 + 1 = 8$
End Points -2 $2\cdot (-2)^{3} - 3 \cdot (-2)^{2} - 12 \cdot (-2) + 1 = -16 - 12 + 24 + 1 = -3$
End Points 3 $2\cdot 3^{3} - 3 \cdot 3^{2} - 12 \cdot 3 + 1 = 54 - 27 -36 + 1 = -8$

Answer

  • Global max value is $y = 8$ attained at $x = -1$ (global max point).
  • Global min value is $y = 19$ attained at $x = 2$ (global min point).
  • Global max is visible as a point on the graph: (-1,8).