Tag: Monotonic

Use derivatives to sketch the graph of a function

by RoRi

October 1, 2015

Let

$f(x) = \frac{1}{6}x^2 + \frac{1}{12} \cos (2x).$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = \frac{1}{3}x - \frac{1}{6} \sin (2x).$

Thus,

$f'(x) = 0 \quad \implies \quad \frac{1}{3} x - \frac{1}{6} \sin (2x) = 0 \quad \implies \quad x = 0.$
$f$ is increasing if $x > 0$ and decreasing if $x < 0$ .
Taking the second derivative,
$f''(x) = \frac{1}{3} - \frac{1}{2} \cos (2x).$

Thus, $f'$ is increasing for all $x$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 30, 2015

Let

$f(x) = x + \cos x.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = 1 - \sin x.$

Thus,

$f'(x) = 0 \quad \implies \quad 1 - \sin x = 0 \quad \implies \quad x = \left(2n + \frac{1}{2}\right) \pi.$
$f$ is increasing for all $x$ since $1 - \sin x \geq 0$ for all $x$ .
Taking the second derivative,
$f''(x) = - \cos x.$

Thus, $f'$ is increasing if $\left( 2n + \frac{1}{2} \right) \pi < x < \left( 2n + \frac{3}{2} \right) \pi$ and is decreasing if $\left( 2n - \frac{1}{2} \right) \pi < x < \left( 2n + \frac{1}{2} \right) \pi$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 30, 2015

Let

$f(x) = x - \sin x.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = 1 - \cos x.$

Thus,

$f'(x) = 0 \quad \implies \quad 1 - \cos x = 0 \quad \implies \quad x = 2n \pi.$
$f$ is increasing for all $x$ since $1 - \cos x \geq 0$ for all $x$ .
Taking the second derivative,
$f''(x) = \sin x.$

Thus, $f'$ is increasing for $2 n \pi < x < (2n+1) \pi$ and decreasing for $(2n-1) \pi < x < 2n \pi$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 30, 2015

Let

$f(x) = \sin^2 x.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = 2 \sin x \cos x = \sin (2x).$

Thus,

$f'(x) = 0 \quad \implies \quad \sin (2x) = 0 \quad \implies \quad x = \frac{n \pi}{2}$

for $n$ an integer. (We found the zeros of sine in this exercise, Apostol Section 2.8 Exercise #1.)
$f$ is increasing if $n \pi < x < \left( n + \frac{1}{2} \right) \pi$ and decreasing if $\left( n - \frac{1}{2} \right) \pi < x < n \pi$ .
Taking the second derivative,
$f''(x) = -2 \cos (2x).$

Thus, $f'$ is increasing if $\left( n - \frac{1}{4} \right) \pi < x < \left( n + \frac{1}{4} \right) \pi$ and decreasing if $\left( n + \frac{1}{4} \right) \pi < x < \left( n + \frac{3}{4} \right) \pi$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 30, 2015

Let

$f(x) = \frac{x^2 - 4}{x^2 - 9}.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = \frac{-10x}{(x^2 - 9)^2}.$

Thus,

$f'(x) = 0 \quad \implies \quad \frac{-10x}{(x^2 -9)^2} = 0 \quad \implies \quad x = 0.$
$f$ is increasing if $x < -3$ or $-3 < x < 0$ and is decreasing if $0 < x < 3$ or $x > 3$ .
Taking the second derivative,
$f''(x) = \frac{30(x^2+3)}{(x^2-9)^3}.$

Thus, $f'$ is increasing for $x < -3$ or $x > 3$ and is decreasing if $-3 < x < 3$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 30, 2015

Let

$f(x) = \frac{x}{1+x^2}.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = \frac{1-x^2}{(1+x^2)^2}.$

Thus,

$f'(x) = 0 \quad \implies \quad \frac{1-x^2}{(1+x^2)^2} = 0 \quad \implies \quad x = \pm 1.$
$f$ is increasing if $|x| < 1$ and decreasing if $|x| > 1$ .
Taking the second derivative,
$f''(x) = \frac{2x(x^2 - 3)}{(1+x^2)^3}.$

Thus, $f'$ is increasing if $-\sqrt{3} < x < 0$ or $x > \sqrt{3}$ and $f'$ is decreasing if $x < - \sqrt{3}$ or $0 < x < \sqrt{3}$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 29, 2015

Let

$f(x) = \frac{1}{(x-1)(x-3)}.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = \frac{-2x + 4}{(x-1)^2 (x-3)^2}.$

Thus,

$f'(x) = 0 \quad \implies \quad \frac{-2x+4}{(x-1)^2(x-3)^2} = 0 \quad \implies \quad x = 2.$
$f$ is increasing if $x < 1$ or $1 < x < 2$ and $f$ is decreasing if $2 < x < 3$ or $x > 3$ . (We have to take some care here to leave out the points $x = 1$ and $x = 3$ since the function is not defined at these points.)
Taking the second derivative,
$f''(x) = \frac{1}{(x-3)^3} - \frac{1}{(x-1)^3}.$

Thus, $f'$ is increasing for $x < 1$ and $x > 3$ and $f'$ is decreasing for $1 < x < 3$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 29, 2015

Let

$f(x) = x + \frac{1}{x^2}.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = 1 - \frac{2}{x^3}.$

Thus,

$f'(x) = 0 \quad \implies \quad 1 - \frac{2}{x^3} = 0 \quad \implies \quad x = 2^{\frac{1}{3}}.$
$f$ is increasing if $x < 0$ or $x > 2^{\frac{1}{3}}$ , and $f$ is decreasing if $0 < x < 2^{\frac{1}{3}}$ .
Taking the second derivative,
$f''(x) = \frac{6}{x^4}.$

Thus, $f'$ is increasing for all $x \neq 0$ (and is undefined if $x = 0$ ).
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 29, 2015

Let

$f(x) = \frac{1}{x^2}.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = - \frac{2}{x^3}.$

Thus,

$f'(x) = 0 \quad \implies \quad -\frac{2}{x^3} = 0 \quad \implies \quad f'(x) \text{ is never zero}.$
$f$ is increasing for $x < 0$ , and decreasing for $x > 0$ .
Taking the second derivative,
$f'(x) = -\frac{2}{x^3} \quad \implies \quad f''(x) = \frac{6}{x^4}.$

Thus, $f'$ is increasing for all $x \neq 0$ (and is undefined at $x = 0$ ).
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 29, 2015

Let

$f(x) = 2 + (x-1)^4.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f'(x) = 4(x-1)^3.$

Thus,

$f'(x) = 0 \quad \implies \quad 4(x-1)^3 = 0 \quad \implies \quad x = 1.$
$f$ is increasing if $x > 1$ and $f$ is decreasing if $x < 1$ .
Taking the second derivative,
$f'(x) = 4(x-1)^3 \quad \implies \quad f''(x) = 12(x-1)^2.$

Thus, $f'$ is increasing for all $x$ since $f''$ is positive for all $x$ .
We sketch the curve,