Tag: Monotonic

Use derivatives to sketch the graph of a function

by RoRi

September 29, 2015

Let

$f(x) = x^3 - 6x^2 + 9x + 5.$

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,
$f'(x) = 3x^2 - 12x + 9.$

Thus,

$f'(x) = 0 \quad \implies \quad 3x^2 - 12x + 9 = 0 \quad \implies \quad x = \{ 1, 3 \}.$
$f$ is increasing for $x < 1$ and $x > 3$ , and decreasing for $1 < x < 3$ .
Taking the second derivative,
$f''(x) = 6x - 12.$

Thus, $f'$ is increasing if $x > 2$ , and decreasing if $x < 2$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 28, 2015

Let

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

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,
$\begin{align*} f'(x) &= 2(x-1)(x+2) + (x-1)^2 \\ &= (x-1)(2x + 4 + x - 1) \\ &= (x-1)(3x + 3) \\ &= 3(x^2 -1) \end{align*}$

Thus,

$f'(x) = 0 \quad \implies \quad 3(x^2-1) = 0 \quad \implies \quad x = \pm 1.$
$f$ is increasing for $|x| > 1$ , and decreasing for $|x| < 1$ .
Taking the second derivative,
$f'(x) = 3(x^2-1) \quad \implies \quad f''(x) = 6x.$

Thus, $f'$ is increasing if $x > 0$ , and decreasing if $x < 0$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 28, 2015

Let

$f(x) = x^3 - 4x.$

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,
$f(x) = x^3 - 4x \quad \implies \quad f'(x) = 3x^2 - 4.$

Thus,

$f'(x) = 0 \quad \implies \quad 3x^2 - 4 = 0 \quad \implies \quad x = \pm \frac{2}{\sqrt{3}}.$
Since $f'(x) > 0$ when $|x| > \frac{2}{\sqrt{3}}$ and $f'(x) < 0$ when $|x| < \frac{2}{\sqrt{3}}$ we have $f$ is increasing if $|x| > \frac{2}{\sqrt{3}}$ and $f$ is decreasing if $|x| < \frac{2}{\sqrt{3}}$ .
Taking the second derivative,
$f'(x) = 3x^2 - 4 \quad \implies \quad f''(x) = 6x.$

Thus, $f'$ is increasing for $x > 0$ ; and decreasing for $x < 0$ .
We sketch the curve,

Use derivatives to sketch the graph of a function

by RoRi

September 28, 2015

Let

$f(x) = x^2 - 3x + 2.$

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,
$f(x) = x^2 - 3x + 2 \quad \implies \quad f'(x) = 2x - 3.$

Thus,

$f'(x) = 0 \quad \implies \quad 2x - 3 = 0 \quad \implies \quad x = \frac{3}{2}.$
Since $f'(x) < 0$ when $x < \frac{3}{2}$ and $f'(x) > 0$ when $x > \frac{3}{2}$ we have $f$ is decreasing if $x < \frac{3}{2}$ and $f$ is increasing if $x > \frac{3}{2}$ .
Taking the second derivative,
$f'(x) = 2x - 3 \quad \implies \quad f''(x) = 2.$

Since this is positive for all $x$ , this means $f'$ is increasing for all $x$ .
We sketch the curve,

Show that a function is monotonic and find a formula for its inverse

by RoRi

September 5, 2015

Let

$f(x) = \begin{cases} x & \text{if } x < 1, \\ x^2 & \text{if } 1 \leq x \leq 4, \\ 8x^{\frac{1}{2}} & \text{if } x > 4. \end{cases}$

Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is strictly increasing on $\mathbb{R}$ we note that it is strictly increasing on each component (since $x^2, \ x,$ and $8x^{1/2}$ are all increasing functions on the domains given). Then we must consider the transition from one of these regions to another. (In other words, we know the function is increasing on each interval, but we need to check that it is increasing from one interval to the next.)

$x_1 < 1 \leq x_2 \quad \implies \quad x_1 < x_2^2 \quad \implies \quad f(x_1) < f(x_2),$

and,

$1 \leq x_1 \leq 4 < x_2 \quad \implies \quad x_1^2 < 8x_2^{\frac{1}{2}} \quad \implies \quad f(x_1) < f(x_2).$

Thus, $f$ is indeed increasing on all of $\mathbb{R}$ .
Next,

$g(y) = \begin{cases} y & \text{if } y < 1, \\ y^{\frac{1}{2}} & \text{if } 1 \leq y \leq 16, \\ \left( \frac{y}{8}\right)^2 & \text{if } y > 16. \end{cases}$

Show a function is monotonic and find a formala for its inverse

by RoRi

September 5, 2015

Let $f(x) = x^3$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then we consider

$\begin{align*} x_2^3 - x_1^3 &= (x_2 - x_1)(x_2^2 + x_1x_2 + x_1^2)\\ &= (x_2 - x_1)\left( \left( x_2 + \frac{x_1}{2} \right)^2 + \frac{3x_1^2}{4} \right). \end{align*}$

But, since $x_2 > x_1$ by assumption, we have $(x_2 - x_1) > 0$ . The second term in the product is also positive since it is a sum of positive terms. Therefore,

$x_2^3 - x_1^3 > 0 \qquad \implies \qquad f(x_2) > f(x_1).$

Hence, $f$ is strictly increasing on $\mathbb{R}$ . (Of course, there are faster ways to discover the $f(x) =x^3$ is increasing, but we do not know yet what is a derivative.)

Next,

$y = x^3 \quad \implies \quad x = y^{\frac{1}{3}} \quad \implies \quad g(y) = y^{\frac{1}{3}} \quad \text{for all } y \in \mathbb{R}.$

Show a function is monotonic and find a formula for its inverse

by RoRi

September 5, 2015

Let $f(x) = 1-x$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then

$x_1 < x_2 \quad \implies \quad -x_1 > -x_2 \quad \implies \quad 1-x_1 > 1-x_2 \quad \implies \quad f(x_1) > f(x_2).$

Hence, $f$ is strictly decreasing on $\mathbb{R}$ .

Next,

$\begin{align*} y = 1-x && \implies && x &= 1-y \\ && \implies && g(y) &= 1-y \qquad \text{for all } y \in \mathbb{R}. \end{align*}$

Show a function is monotonic and find a formula for its inverse

by RoRi

September 5, 2015

Let $f(x) = 2x+5$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then

$\begin{align*} x_1 < x_2 && \implies && 2x_1 < 2x_2 \\ && \implies && 2x_1 + 5 < 2x_2 + 5 \\ && \implies && f(x_1) < f(x_2). \end{align*}$

Hence, $f$ is strictly increasing on $\mathbb{R}$ .

Next,

$\begin{align*} y = 2x + 5 && \implies && x &= \frac{y-5}{2} \\ && \implies && g(y) &= \frac{y-5}{2} \qquad \text{for all } y \in \mathbb{R}. \end{align*}$

Show a function is monotonic and find a formula for its inverse

by RoRi

September 5, 2015

Let $f(x) = x + 1$ . Show that $f$ is strictly monotonic on $\mathbb{R}$ . Find the domain of the inverse of $f$ , denoted by $g$ . Find a formula for computing $g(y)$ for each $y$ in the domain of $g$ .

First, to show $f$ is monotonic let $x_1, x_2 \in \mathbb{R}$ with $x_1 < x_2$ . Then

$x_1 + 1 < x_2 + 1 \quad \implies \quad f(x_1) < f(x_2).$

Hence, $f$ is strictly increasing on $\mathbb{R}$ .

Next,

$y = x+1 \quad \implies \quad x = y-1 \quad \implies \quad g(y) = y-1 \quad \text{for all } y \in \mathbb{R}.$