Chapter 6, Problem 1RE

(a)

To determine

The value of $\left(f\circ g\right)\left(2\right)$ for the functions $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

Answer to Problem 1RE

The value of the function $\left(f\circ g\right)\left(x\right)$ at $x=2$ is $\underset{\_}{-26}$.

Explanation of Solution

Definition used:

Given two functions f and g, the composite function, denoted by $f\circ g$ (read as “f composed with g”), is defined by $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$.

Calculation:

Given that $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

By using the above definition $\left(f\circ g\right)\left(2\right)$ can be calculated as follows.

$\begin{array}{c}\left(f\circ g\right)\left(2\right)=f\left(g\left(2\right)\right)\\ =f\left(1-2{\left(2\right)}^2\right)\left[\because g\left(x\right)=1-2x^2\right]\\ =f\left(-7\right)\\ =3\left(-7\right)-\text{5 }\left[\because f\left(x\right)=3x-5\right]\\ =-26\end{array}$

Thus, the value of the function $\left(f\circ g\right)\left(x\right)$ at $x=2$ is $\underset{\_}{-26}$.

(b)

To determine

The value of $\left(g\circ f\right)\left(-2\right)$ for the functions $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

Answer to Problem 1RE

The value of the function $\left(g\circ f\right)\left(x\right)$ at $x=-2$ is $\underset{\_}{-241}$.

Explanation of Solution

Given that $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

By using the above definition $\left(g\circ f\right)\left(-2\right)$ can be calculated as follows.

$\begin{array}{c}\left(g\circ f\right)\left(-2\right)=g\left(f\left(-2\right)\right)\\ =g\left(3\left(-2\right)-5\right)\left[\because f\left(x\right)=3x-5\right]\\ =g\left(-11\right)\\ =1-2{\left(-11\right)}^2\left[\because g\left(x\right)=1-2x^2\right]\\ =-241\end{array}$

Thus, the value of the function $\left(g\circ f\right)\left(x\right)$ at $x=-2$ is $\underset{\_}{-241}$.

(c)

To determine

The value of $\left(f\circ f\right)\left(4\right)$ for the functions $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

Answer to Problem 1RE

The value of the function $\left(f\circ f\right)\left(x\right)$ at $x=4$ is $\underset{\_}{16}$.

Explanation of Solution

Given that $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

By using the above definition $\left(f\circ f\right)\left(4\right)$ can be calculated as follows.

$\begin{array}{c}\left(f\circ f\right)\left(4\right)=f\left(f\left(4\right)\right)\\ =f\left(3\left(4\right)-5\right)\left[\because f\left(x\right)=3x-5\right]\\ =f\left(7\right)\\ =3\left(7\right)-5\left[\because f\left(x\right)=3x-5\right]\\ =16\end{array}$

Thus, the value of the function $\left(f\circ f\right)\left(x\right)$ at $x=4$ is $\underset{\_}{16}$.

(d)

To determine

The value of $\left(g\circ g\right)\left(-1\right)$ for the functions $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

Answer to Problem 1RE

The value of the function $\left(g\circ g\right)\left(x\right)$ at $x=-1$ is $\underset{\_}{-1}$.

Explanation of Solution

Given that $f\left(x\right)=3x-5$ and $g\left(x\right)=1-2x^2$.

By using the above definition $\left(g\circ g\right)\left(-1\right)$ can be calculated as follows.

$\begin{array}{c}\left(g\circ g\right)\left(-1\right)=g\left(g\left(-1\right)\right)\\ =g\left(1-2{\left(-1\right)}^2\right)\left[\because g\left(x\right)=1-2x^2\right]\\ =g\left(-1\right)\\ =1-2{\left(-1\right)}^2\left[\because g\left(x\right)=1-2x^2\right]\\ =-1\end{array}$

Thus, the value of the function $\left(g\circ g\right)\left(x\right)$ at $x=-1$ is $\underset{\_}{-1}$.