Math  /  Algebra

Question9. f:x2x1g:x3x,x0\begin{array}{l} f: x \mapsto 2 x-1 \\ g: x \mapsto \frac{3}{x}, x \neq 0 \end{array} May O4 3H (a) Find the value of (i) f(3)\mathrm{f}(\mathrm{3}), (ii) fg(6)\mathrm{fg}(6). (b) Express the inverse function f1\mathrm{f}^{-1} in the form f1:x\mathrm{f}^{-1}: x \mapsto \ldots (c) (i) Express the composite function gf in the form gf:x\mathrm{gf}: x \mapsto \ldots. (ii) Which value of xx must be excluded from the domain of gg ?

Studdy Solution

STEP 1

What is this asking? We're given two functions, f(x)f(x) and g(x)g(x), and we need to evaluate them at specific points, find the inverse of f(x)f(x), and figure out the composite function g(f(x))g(f(x)). Watch out! Don't mix up the order of operations when composing functions!
Also, remember to consider any values that might make the denominator zero.

STEP 2

1. Evaluate f(3)
2. Evaluate fg(6)
3. Find the inverse of f(x)
4. Find the composite function gf(x)
5. Determine the excluded value

STEP 3

We're asked to **evaluate** f(3)f(3).
Remember, this just means plugging x=3x = \textbf{3} into the function f(x)=2x1f(x) = 2x - 1.

STEP 4

f(3)=2(3)1f(3) = 2 \cdot (\textbf{3}) - 1

STEP 5

f(3)=61f(3) = \textbf{6} - 1

STEP 6

f(3)=5f(3) = \textbf{5} So, f(3)f(3) equals **5**!

STEP 7

Now, let's **tackle** fg(6)fg(6).
This means we first evaluate g(6)g(6) and *then* plug that result into f(x)f(x).

STEP 8

First, let's find g(6)g(6).
We have g(x)=3xg(x) = \frac{3}{x}, so: g(6)=36g(6) = \frac{3}{\textbf{6}}

STEP 9

We can simplify the fraction by dividing the numerator and denominator by their **greatest common divisor**, which is 3. g(6)=3÷36÷3=12g(6) = \frac{3 \div 3}{6 \div 3} = \frac{1}{2} So, g(6)=12g(6) = \frac{\textbf{1}}{\textbf{2}}.

STEP 10

Now, we plug this result into f(x)f(x): f(g(6))=f(12)=2(12)1f(g(6)) = f\left(\frac{1}{2}\right) = 2 \cdot \left(\frac{\textbf{1}}{\textbf{2}}\right) - 1

STEP 11

Multiplying 2 by 12\frac{1}{2} gives us 1. f(12)=11f\left(\frac{1}{2}\right) = \textbf{1} - 1

STEP 12

f(12)=0f\left(\frac{1}{2}\right) = \textbf{0} Therefore, fg(6)=0fg(6) = \textbf{0}!

STEP 13

To find the **inverse** of f(x)f(x), we **replace** f(x)f(x) with yy, **swap** xx and yy, and then **solve** for yy.

STEP 14

We have f(x)=2x1f(x) = 2x - 1, so we start with y=2x1y = 2x - 1.
Swapping xx and yy gives us x=2y1x = 2y - 1.

STEP 15

Now, let's **isolate** yy.
We can add 1 to both sides of the equation: x+1=2yx + 1 = 2y

STEP 16

Next, we divide both sides by 2: x+12=2y2\frac{x+1}{2} = \frac{2y}{2} x+12=y\frac{x+1}{2} = y

STEP 17

So, the inverse function is f1(x)=x+12f^{-1}(x) = \frac{x+1}{2}.

STEP 18

To find gf(x)gf(x), we **substitute** f(x)f(x) into g(x)g(x).

STEP 19

We have f(x)=2x1f(x) = 2x - 1 and g(x)=3xg(x) = \frac{3}{x}.
So, gf(x)=g(f(x))=g(2x1)gf(x) = g(f(x)) = g(2x - 1).

STEP 20

Substituting 2x12x - 1 into g(x)g(x) gives us: gf(x)=32x1gf(x) = \frac{3}{2x - 1}

STEP 21

We need to find the value of xx that makes the denominator of gf(x)gf(x) equal to zero, since division by zero is undefined.

STEP 22

We set the denominator equal to zero and solve for xx: 2x1=02x - 1 = 0

STEP 23

Adding 1 to both sides gives us: 2x=12x = 1

STEP 24

Dividing both sides by 2 gives us: x=12x = \frac{1}{2}

STEP 25

(a) (i) f(3)=5f(3) = 5 (ii) fg(6)=0fg(6) = 0 (b) f1(x)=x+12f^{-1}(x) = \frac{x+1}{2} (c) (i) gf(x)=32x1gf(x) = \frac{3}{2x - 1} (ii) x=12x = \frac{1}{2} is excluded from the domain of gf(x)gf(x).

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