Tag: Composition

Compute values of a composite function given values of the components

by RoRi

September 22, 2015

Given functions $f(x)$ and $g(x)$ with the following values

$\begin{tabular}{|c | c | c | c | c |} \hline $x$ & $f(x)$ & $f'(x)$ & $g(x)$ & $g'(x)$ \\ \hline 0 & 1 & 5 & 2 & -5 \\ 1 & 3 & -2 & 0 & 1 \\ 2 & 0 & 2 & 3 & 1 \\ 3 & 2 & 4 & 1 & -6 \\ \hline \end{tabular}$

Let $h(x) = f(g(x))$ and $k(x) = g(f(x))$ and construct a table as above for the functions $h$ and $k$ .

First, by the chain rule we have

$\begin{align*} h(x) &= f(g(x)) &\implies \quad h'(x) &= f'(g(x)) g'(x) \\ k(x) &= g(f(x)) &\implies \quad k'(x) &= g'(f(x)) f'(x). \end{align*}$

Using these formulas we compute the table,

$\begin{tabular}{|c | c | c | c | c |} \hline $x$ & $h(x)$ & $h'(x)$ & $k(x)$ & $k'(x)$ \\ \hline 0 & 0 & -10 & 0 & 5 \\ 1 & 1 & 5 & 1 & 12 \\ 2 & 2 & 4 & 2 & -10 \\ 3 & 3 & 12 & 3 & 4 \\ \hline \end{tabular}$

Find points at which a given composite function is continuous

by RoRi

September 4, 2015

Define functions $f$ and $g$ as follows:

$f(x) = \frac{x+|x|}{2}, \text{ for all } x, \qquad g(x) = \begin{cases} x & \text{for } x < 0,\\ x^2 &\text{for } x \geq 0. \end{cases}$

Find a formula for $h(x) = g(f(x))$ . Find the values at which $h(x)$ is continuous.

If $x < 0$ , we have $|x| = -x$ and we compute the composition,

$h(x) = g(f(x)) = g \left( \frac{x - x}{2} \right) = g(0) = 0.$

If $x \geq 0$ , then $|x| = x$ and we compute the composition,

$h(x) = g(f(x)) = g \left( \frac{x+x}{2} \right) = g(x) = x^2.$

Hence,

$h(x) = \begin{cases} 0 & \text{if } x < 0, \\ x^2 & \text{if } x \geq 0. \end{cases}$

This is the same function as in this previous exercise, and it is continuous everywhere.

Find points at which a given composite function is continuous

by RoRi

September 3, 2015

Define functions $f$ and $g$ as follows:

$f(x) = \begin{cases} 1 & \text{if } |x| \leq 1 \\ 0 & \text{if } |x| > 1. \end{cases}, \qquad g(x) = \begin{cases} 2-x^2 & \text{if } |x| \leq 2 \\ 2 & \text{if } |x| > 2. \end{cases}$

Find a formula for $h(x) = f(g(x))$ and find the values at which $h(x)$ is continuous.

If $|x| > 2$ then $g(x) = 2$ and so $|g(x)| > 1$ . Therefore, by the definition of $f$ ,

$h(x) = f(g(x)) = 0.$

If $\sqrt{3} < x \leq 2$ , then $|g(x)| = |2 - x^2| > 1$ and so again

$h(x) = f(g(x)) = 0.$

Next, if $1 \leq x \leq \sqrt{3}$ then $|g(x)| = | 2 - x^2| \leq 1$ and so

$h(x) = f(g(x)) = 1.$

Finally, if $|x| < 1$ , then $|g(x)| = |2 - x^2| > 1$ and we have

$h(x) = f(g(x)) = 0.$

Putting this all together we have

$h(x) = \begin{cases} 1 & \text{if } 1 \leq |x| \leq \sqrt{3} \\ 0 & \text{otherwise}. \end{cases}$

From this expression we have $h(x)$ is continuous everywhere except at $|x| = 1$ and $|x| = \sqrt{3}$ .

Find points at which a given composite function is continuous

by RoRi

September 3, 2015

Define functions $f$ and $g$ as follows:

$f(x) = \frac{x+|x|}{2}, \text{ for all } x, \qquad g(x) = \begin{cases} x & \text{for } x < 0,\\ x^2 &\text{for } x \geq 0. \end{cases}$

Find a formula for $h(x) = f(g(x))$ . Find the values at which $h(x)$ is continuous.

If $x < 0$ , we have $|x| = -x$ and we compute the composition,

$h(x) = f(g(x)) = f(x) = \frac{x-x}{2} = 0.$

If $x \geq 0$ , then $|x| = x$ and we compute the composition,

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

Putting these together,

$h(x) = \begin{cases} 0 & \text{if } x < 0, \\ x^2 & \text{if } x \geq 0. \end{cases}$

Certainly $h$ is continuous for all $x \neq 0$ since the constant function $h(x) =0$ and the polynomial $h(x) = x^2$ are continuous. Further, $h$ is continuous at $x = 0$ since

$\lim_{x \to 0^-} h(x) = \lim_{x \to 0^+} h(x) = \lim_{x \to 0} h(x) = 0 = h(0).$

Find a formula for the composition of given functions

by RoRi

September 1, 2015

Find the composition function $h(x) = (f \circ g)(x) = f(g(x))$ of the functions:

$f(x) = \sqrt{x+\sqrt{x}}, \qquad \text{if } x > 0, \qquad g(x) = x + \sqrt{x} \quad \text{if } x > 0.$

We compute $h(x) = f(g(x))$ ,

$\begin{align*} h(x) &= f(g(x)) \\ &= f(x + \sqrt{x}) \\ &= \sqrt{x + \sqrt{x} + \sqrt{x + \sqrt{x}}} \end{align*}$

This is valid for all $x > 0$ . (Of course, this formula would generally be valid for $x = 0$ as well, but since the domain of $f$ and $g$ is given as $x > 0$ , the composition is defined only for $x>0$ .)

Find a formula for the composition of given functions

by RoRi

September 1, 2015

Find the composition function $h(x) = (f \circ g)(x) = f(g(x))$ of the functions:

$f(x) = \sqrt{x}, \qquad \text{if } x > 0, \qquad g(x) = x + \sqrt{x} \quad \text{if } x > 0.$

We compute $h(x) = f(g(x))$ ,

$\begin{align*} h(x) &= f(g(x)) \\ &= f(x + \sqrt{x}) \\ &= \sqrt{ x + \sqrt{x}} \\ \end{align*}$

Find a formula for the composition of given functions

by RoRi

September 1, 2015

Find the composition function $h(x) = (f \circ g)(x) = f(g(x))$ of the functions:

$f(x) = \sqrt{x}, \qquad \text{if } x \geq 0, \qquad g(x) = \sin x.$

We compute $h(x) = f(g(x))$ ,

$\begin{align*} h(x) &= f(g(x)) \\ &= f(\sin x) \\ &= \sqrt{\sin x} \end{align*}$

This is valid for all $2n \pi \leq x \leq (2n+1) \pi$ where $n \in \mathbb{Z}$ . (These are the values for which $\sin x \geq 0$ .)

Find a formula for the composition of given functions

by RoRi

August 31, 2015

Find the composition function $h(x) = (f \circ g)(x) = f(g(x))$ of the functions:

$f(x) = \sin x, \qquad g(x) = \sqrt{x} \quad \text{if } x \geq 0.$

We compute $h(x) = f(g(x))$ ,

$\begin{align*} h(x) &= f(g(x)) \\ &= f(\sqrt{x}) \\ &= \sin \sqrt{x}. \end{align*}$

This is valid for all $x \geq 0$ .

Find a formula for the composition of given functions

by RoRi

August 31, 2015

Find the composition function $h(x) = (f \circ g)(x) = f(g(x))$ of the functions:

$f(x) = - x^2, \qquad g(x) = \sqrt{x} \quad \text{if } x \geq 0.$

We compute $h(x) = f(g(x))$ ,

$\begin{align*} h(x) &= f(g(x)) \\ &= f(\sqrt{x}) \\ &= - \left( \sqrt{x} \right)^2 \\ &= -x. \end{align*}$

This is valid for all $x \geq 0$ .

Find a formula for the composition of given functions

by RoRi

August 31, 2015

Find the composition function $h(x) = (f \circ g)(x) = f(g(x))$ of the functions:

$f(x) = x^2, \qquad g(x) = \sqrt{x} \quad \text{if } x \geq 0.$

We compute $h(x) = f(g(x))$ ,

$\begin{align*} h(x) &= f(g(x)) \\ &= f(\sqrt{x}) \\ &= \left( \sqrt{x} \right)^2 \\ &= x \end{align*}$

This is valid for all $x \geq 0$ .