Monotonicity

Evaluation

To find the intervals of monotonicity of $f$ :

Find the domain of $f$ .
Find and simplify $f^{'}$ .
Find all critical points of $f$ .
Construct a sign table with the critical points.

Examples

Find the intervals of monotonicity, critical points and local extrema of $f (x) = \frac{x^{2} + 3}{x - 1}$ .

$f (x)$ is undefined when $x - 1 \neq 0 ⟹ x \neq 1$ . So, the domain of $f$ is $x \in R ∖ {1}$ .

\begin{aligned} f^{'} (x) & = \frac{d}{d x} \frac{x^{2} + 3}{x - 1} \\ = \frac{(x - 1) 2 x - (x^{2} + 3)}{(x - 1)^{2}} \\ = \frac{(x - 3) (x + 1)}{(x - 1)^{2}} \\ ⟹ 0 & = x - 3 or 0 = x + 1 \\ x & = 3 or x = - 1 \end{aligned}

$x = - 1$ and $x = 3$ are critical points. Since $x = 1 \notin d o m (f)$ , it cannot be a critical point.

$- \infty$		-1		1		3		$\infty$
$f^{'} (x)$	+		-		-		+
$f (x)$	inc		dec		dec		inc

$f$ has a relative maximum at $x = - 1$ and a relative minimum at $x = 3$ .
$x = 1$ cannot be a local extremum since $x = 1 \notin d o m (f)$ .

Find the intervals of monotonicity, critical points and local extrema of

f (x) = x^{2} (x^{2} - 1)^{1 / 3}

given

f^{'} (x) = \frac{2 x (4 x^{2} - 3)}{3 (x^{2} - 1)^{2 / 3}}

$f^{'} (x)$ is defined $\forall x \in R$ .

\begin{aligned} f^{'} (x) & = \frac{2 x (4 x^{2} - 3)}{3 (x^{2} - 1)^{2 / 3}} \\ ⟹ 0 & = 2 x or 0 = 4 x^{2} - 3 \\ x & = 0 or \pm \frac{\sqrt{3}}{2} \end{aligned}

$f^{'} (x)$ is undefined when $3 (x^{2} - 1)^{2 / 3} = 0 ⟹ x = \pm 1$ .
$x = 0$ , $x = \pm \frac{\sqrt{3}}{2}$ and $x = \pm 1$ are critical points.

$- \infty$		-1		$- \frac{\sqrt{3}}{2}$		0		$\frac{\sqrt{3}}{2}$		1		$\infty$
$f^{'} (x)$	-		-		+		-		+		+
$f (x)$	dec		dec		inc		dec		inc		inc

At $x = \pm 1$ , this is no extremum since $f^{'}$ does not change sign.
At $x = \pm \frac{\sqrt{3}}{2}$ , $f$ has a local minimum, and at $x = 0$ there is a local maximum.

Contributors

Ajitani Hifumi

Changelog

Last edited 29 days ago

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Monotonicity ​

Evaluation ​

Examples ​

Contributors

Changelog

Monotonicity

Evaluation

Examples