The problem defines two functions, $f(x) = x^2 - 1$ and $g(x) = 3x + 2$. Part (a) asks us to simplify three expressions: (i) $2f - 3g$, (ii) $g(f(x))$, and (iii) $g^{-1}(f(x))$. Part (b) asks to show that $f(x)$ is not one-to-one, but $g(x)$ is one-to-one.

1. Problem Description

The problem defines two functions, $f(x) = x^2 - 1$ and $g(x) = 3x + 2$. Part (a) asks us to simplify three expressions: (i) $2f - 3g$, (ii) $g(f(x))$, and (iii) $g^{-1}(f(x))$. Part (b) asks to show that $f(x)$ is not one-to-one, but $g(x)$ is one-to-one.

2. Solution Steps

(a) Simplify

(i) $2f - 3g$

First, we write down the expressions for $2f(x)$ and $3g(x)$.

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

$3g(x) = 3(3x + 2) = 9x + 6$

Then,

$2f(x) - 3g(x) = (2x^2 - 2) - (9x + 6) = 2x^2 - 2 - 9x - 6 = 2x^2 - 9x - 8$

(ii) $g(f(x))$

We need to find the composite function $g(f(x))$. Since $f(x) = x^2 - 1$ and $g(x) = 3x + 2$, we have

$g(f(x)) = g(x^2 - 1) = 3(x^2 - 1) + 2 = 3x^2 - 3 + 2 = 3x^2 - 1$

(iii) $g^{-1}(f(x))$

First, we need to find the inverse function $g^{-1}(x)$. Since $g(x) = 3x + 2$, we set $y = 3x + 2$ and solve for $x$ in terms of $y$:

$y = 3x + 2$

$y - 2 = 3x$

$x = \frac{y - 2}{3}$

So, $g^{-1}(x) = \frac{x - 2}{3}$.

Then, we can find $g^{-1}(f(x))$:

$g^{-1}(f(x)) = g^{-1}(x^2 - 1) = \frac{(x^2 - 1) - 2}{3} = \frac{x^2 - 3}{3}$

(b) One-to-one functions

To show that $f(x)$ is not one-to-one, we need to find two different values of $x$, say $x_1$ and $x_2$, such that $f(x_1) = f(x_2)$.

Let $x_1 = 0$ and $x_2 = 0$.

$f(0) = 0^2 - 1 = -1$

$f(0) = (-0)^2 - 1 = -1$

Since $f(0) = f(0)$ but $0 \neq -0$ is false, however if we take $x_1=2$ and $x_2=-2$, $f(2) = 2^2-1 = 3$ and $f(-2) = (-2)^2-1 = 3$. So $f(2) = f(-2)$ and $2 \neq -2$, so $f(x)$ is not one-to-one.

To show that $g(x)$ is one-to-one, we need to show that if $g(x_1) = g(x_2)$, then $x_1 = x_2$.

Suppose $g(x_1) = g(x_2)$. Then $3x_1 + 2 = 3x_2 + 2$.

Subtracting 2 from both sides, we get $3x_1 = 3x_2$.

Dividing both sides by 3, we get $x_1 = x_2$.

Therefore, $g(x)$ is one-to-one.

3. Final Answer

(a)

(i) $2f(x) - 3g(x) = 2x^2 - 9x - 8$

(ii) $g(f(x)) = 3x^2 - 1$

(iii) $g^{-1}(f(x)) = \frac{x^2 - 3}{3}$

(b)

$f(x)$ is not one-to-one because $f(2) = f(-2)$ but $2 \neq -2$.

$g(x)$ is one-to-one because if $g(x_1) = g(x_2)$, then $x_1 = x_2$.