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.

AlgebraFunctionsFunction CompositionInverse FunctionsOne-to-one Functions
2025/3/25

1. Problem Description

The problem defines two functions, f(x)=x21f(x) = x^2 - 1 and g(x)=3x+2g(x) = 3x + 2. Part (a) asks us to simplify three expressions: (i) 2f3g2f - 3g, (ii) g(f(x))g(f(x)), and (iii) g1(f(x))g^{-1}(f(x)). Part (b) asks to show that f(x)f(x) is not one-to-one, but g(x)g(x) is one-to-one.

2. Solution Steps

(a) Simplify
(i) 2f3g2f - 3g
First, we write down the expressions for 2f(x)2f(x) and 3g(x)3g(x).
2f(x)=2(x21)=2x222f(x) = 2(x^2 - 1) = 2x^2 - 2
3g(x)=3(3x+2)=9x+63g(x) = 3(3x + 2) = 9x + 6
Then,
2f(x)3g(x)=(2x22)(9x+6)=2x229x6=2x29x82f(x) - 3g(x) = (2x^2 - 2) - (9x + 6) = 2x^2 - 2 - 9x - 6 = 2x^2 - 9x - 8
(ii) g(f(x))g(f(x))
We need to find the composite function g(f(x))g(f(x)). Since f(x)=x21f(x) = x^2 - 1 and g(x)=3x+2g(x) = 3x + 2, we have
g(f(x))=g(x21)=3(x21)+2=3x23+2=3x21g(f(x)) = g(x^2 - 1) = 3(x^2 - 1) + 2 = 3x^2 - 3 + 2 = 3x^2 - 1
(iii) g1(f(x))g^{-1}(f(x))
First, we need to find the inverse function g1(x)g^{-1}(x). Since g(x)=3x+2g(x) = 3x + 2, we set y=3x+2y = 3x + 2 and solve for xx in terms of yy:
y=3x+2y = 3x + 2
y2=3xy - 2 = 3x
x=y23x = \frac{y - 2}{3}
So, g1(x)=x23g^{-1}(x) = \frac{x - 2}{3}.
Then, we can find g1(f(x))g^{-1}(f(x)):
g1(f(x))=g1(x21)=(x21)23=x233g^{-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)f(x) is not one-to-one, we need to find two different values of xx, say x1x_1 and x2x_2, such that f(x1)=f(x2)f(x_1) = f(x_2).
Let x1=0x_1 = 0 and x2=0x_2 = 0.
f(0)=021=1f(0) = 0^2 - 1 = -1
f(0)=(0)21=1f(0) = (-0)^2 - 1 = -1
Since f(0)=f(0)f(0) = f(0) but 000 \neq -0 is false, however if we take x1=2x_1=2 and x2=2x_2=-2, f(2)=221=3f(2) = 2^2-1 = 3 and f(2)=(2)21=3f(-2) = (-2)^2-1 = 3. So f(2)=f(2)f(2) = f(-2) and 222 \neq -2, so f(x)f(x) is not one-to-one.
To show that g(x)g(x) is one-to-one, we need to show that if g(x1)=g(x2)g(x_1) = g(x_2), then x1=x2x_1 = x_2.
Suppose g(x1)=g(x2)g(x_1) = g(x_2). Then 3x1+2=3x2+23x_1 + 2 = 3x_2 + 2.
Subtracting 2 from both sides, we get 3x1=3x23x_1 = 3x_2.
Dividing both sides by 3, we get x1=x2x_1 = x_2.
Therefore, g(x)g(x) is one-to-one.

3. Final Answer

(a)
(i) 2f(x)3g(x)=2x29x82f(x) - 3g(x) = 2x^2 - 9x - 8
(ii) g(f(x))=3x21g(f(x)) = 3x^2 - 1
(iii) g1(f(x))=x233g^{-1}(f(x)) = \frac{x^2 - 3}{3}
(b)
f(x)f(x) is not one-to-one because f(2)=f(2)f(2) = f(-2) but 222 \neq -2.
g(x)g(x) is one-to-one because if g(x1)=g(x2)g(x_1) = g(x_2), then x1=x2x_1 = x_2.

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