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Using derivative to study monotonicity
What we know about the relation between $f^{\prime}$ and $f$.
Extreme Value Theorem: If $c$ is a local extreme point, then $f^{\prime}(c)= 0.$ (if defined).
Corollary: If $f^{\prime}(x) = 0$ on interval, then $f$ is constant.
This is helpful but what about increasing/decreasing behavior?
It is determined by the sign of $f^{\prime}(x)$!
Example. Consider the general linear function: $$\begin{aligned} f(x) &= mx + b \\ f^{\prime}(x) & = m \end{aligned}$$ Then $m$ determines what happens to $f$.
$$\begin{aligned} m > 0 & \Rightarrow f \text{ increasing} \\ m < 0 & \Rightarrow f \text{ decreasing} \\ m = 0 & \Rightarrow f \text{ constant} \end{aligned}$$
Generalize this to all functions...
Monotonicity Theorem.
- If $f^{\prime}(x) > 0$ on an open interval, then $f$ is increasing on this interval.
- If $f^{\prime}(x) < 0$ on an open interval, then $f$ is decreasing on this interval.
Example. Given $$f(x) = x^{3} - 3x, $$ on what intervals is it increasing and decreasing? $$f^{\prime}(x) = 3x^{2} - 3$$ Find $x$'s such $f^{\prime}(x) > 0$ or $f^{\prime}(x) < 0$. Solve these inequalities.
Start with this equation $f^{\prime}(x) = 0$. $$\begin{aligned} 3x^{2} - 3 &= 0 \\ x^{2} - 1 &= 0 \\ x^{2} &= 1 \\ x & = \pm 1 \end{aligned}$$
We have three intervals
- $(-\infty, -1),$
- $(-1,1),$
- $(1,\infty).$
We need to know the sign of the derivative on each.
We can assume that the sign of $f^{\prime}$ changes only at $-1,1$. Then look any point within each interval to answer:
- Pick $x = -2$, $f^{\prime}(-2) = 3 \cdot (-2)^{2} -3 = 9 > 0$
- then
- $$f^{\prime}(x) > 0 \forall x \text{ in } (-\infty, -1).$$
- Pick $x = 0$, $f^{\prime}(0) = -3 < 0$ so
- $$ f^{\prime}(x) <0 \forall x \text{ in } (-1,1).$$
- Pick $x=2$, $f^{\prime}(2) = 3 \cdot 2^{2} - 3 = 9 > 0$
- then
- $$ f^{\prime}(x) >0 \forall x \text{ in } (1,\infty).$$
So
- $f \nearrow$ on $(-\infty,-1)$.
- $f \searrow$ on $(-1,1)$.
- $f \nearrow$ on $(1,\infty)$.
Corollary. When two have the same derivative, you can get the graph of one from the other by a vertical shift.
More on the example, what do we know about $\pm 1$? The derivative is 0 so these may be extreme points.
They are. But we know more! Specifically,
- -1 is a local max point and
- 1 is a local min point.
How do we know?
Consider the results of analysis above.
- $f' > 0$: $f\nearrow$ to the left of -1,
- $f' < 0$: $f\searrow$ to the right of -1.
$\Leftrightarrow -1$ is a local max. The opposite for $x = 1$.
More generally:
Theorem: First Derivative Test. Suppose $c$ is a critical point of $f$. Then
- if $f^{\prime}(x)$ changes its sign from + to -, then $c$ is a local max point;
- if $f^{\prime}(x)$ changes its sign from – to +, this is a local min point.
Proof. Follows from the Monotonicity Theorem. $\blacksquare$
Proof of Monotonicity Theorem. We use the Mean Value Theorem. We need to show that if $a<b$ then $f(a) < f(b)$ (i.e., it's increasing). Recall MVT: $$\frac{f(b)-f(a)}{b-a} = f^{\prime}(c)$$ for some $c$ in $(a,b)$.
But, the theorem assumes $f^{\prime}(x) > 0$ for all $x$, so it turns into: $$ f^{\prime}(c) > 0$$ where $f^{\prime}(c)$ is the slope of the secant line.
Some algebra : $$\frac{f(b) – f(a)}{b-a} > 0.$$ Now observe:
- denominator $b-a > 0$,
- so the numerator > 0, or
- $f(b) – f(a) > 0$, or
- $f(b) > f(a).$
$\blacksquare$
Example. Consider $$\begin{alignat}{3} f(x) &= \sin x & \quad & \nearrow & \quad & \searrow \\ f^{\prime}(x) &= \cos x & \quad & + & \quad & - \end{alignat}$$
Need to solve the inequalities: $\cos x > 0, \cos x < 0$.
Start with with equations: $\cos x = 0, \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$.