To find the compositions $ fg $ and $ gf $, we need to understand what these notations mean.

1. **Composition of Functions**: 
   - $ fg $ means $ f(g(x)) $
   - $ gf $ means $ g(f(x)) $

Given the functions:
- $ f(x) = \frac{1}{z-2} $ where $ z = 2 $ (this seems to be a constant function since $ z $ is always 2, making $ f(x) $ undefined)
- $ g(x) = 6x + 1 $

### Step 1: Analyze $ f(x) $

Since $ z = 2 $, we can see that $ f(x) $ is undefined because it leads to division by zero:
\[
f(x) = \frac{1}{2-2} = \frac{1}{0}
\]
Thus, $ f(x) $ does not produce a valid output for any input $ x $.

### Step 2: Calculate $ fg $

Now, let's calculate $ fg $:
\[
fg = f(g(x)) = f(6x + 1)
\]
Since $ f(x) $ is undefined, $ fg $ will also be undefined.

### Step 3: Calculate $ gf $

Next, we calculate $ gf $:
\[
gf = g(f(x)) = g\left(\frac{1}{z-2}\right)
\]
Again, since $ f(x) $ is undefined, $ gf $ will also be undefined.

### Conclusion

Both compositions $ fg $ and $ gf $ are undefined due to the nature of the function $ f(x) $.

Thus, the final answers are:
- (ii) $ fg $ is undefined
- (iii) $ gf $ is undefined

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Question

$(c) $ f: x \rightarrow \frac{1}{z-2}, z=2 $ $ g: x \rightarrow 6 x+1 $ Cari/ Find: (ii) fg (iii) $ g f $$

(c) $ f: x \rightarrow \frac{1}{z-2}, z=2 $ $ g: x \rightarrow 6 x+1 $ Cari/ Find: (ii) fg (iii) $ g f $

Ask by Bird Higgins.

Mar 19,2025 11:09

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Tutor-Verified Answer

Answer

$ fg $ and $ gf $ are both undefined because $ f(x) $ is undefined.

Solution

To find the compositions $ fg $ and $ gf $, we need to understand what these notations mean. 1. **Composition of Functions**: - $ fg $ means $ f(g(x)) $ - $ gf $ means $ g(f(x)) $ Given the functions: - $ f(x) = \frac{1}{z-2} $ where $ z = 2 $ (this seems to be a constant function since $ z $ is always 2, making $ f(x) $ undefined) - $ g(x) = 6x + 1 $ ### Step 1: Analyze $ f(x) $ Since $ z = 2 $, we can see that $ f(x) $ is undefined because it leads to division by zero: \[ f(x) = \frac{1}{2-2} = \frac{1}{0} \] Thus, $ f(x) $ does not produce a valid output for any input $ x $. ### Step 2: Calculate $ fg $ Now, let's calculate $ fg $: \[ fg = f(g(x)) = f(6x + 1) \] Since $ f(x) $ is undefined, $ fg $ will also be undefined. ### Step 3: Calculate $ gf $ Next, we calculate $ gf $: \[ gf = g(f(x)) = g\left(\frac{1}{z-2}\right) \] Again, since $ f(x) $ is undefined, $ gf $ will also be undefined. ### Conclusion Both compositions $ fg $ and $ gf $ are undefined due to the nature of the function $ f(x) $. Thus, the final answers are: - (ii) $ fg $ is undefined - (iii) $ gf $ is undefined

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The Deep Dive

To find $ fg $ and $ gf $, we need to understand the compositions of the functions. Given: $ f(z) = \frac{1}{z - 2} $ For $ g(x) = 6x + 1 $, we need to express $ fg $ and $ gf $: (ii) For $ fg(x) $: First, we substitute $ g(x) $ into $ f(z) $: \[ fg(x) = f(g(x)) = f(6x + 1) = \frac{1}{(6x + 1) - 2} = \frac{1}{6x - 1} \] (iii) For $ gf(x) $: Next, substitute $ f(z) $ into $ g(x) $: \[ gf(z) = g(f(z)) = g\left(\frac{1}{z - 2}\right) = 6\left(\frac{1}{z - 2}\right) + 1 = \frac{6}{z - 2} + 1 = \frac{6 + (z - 2)}{z - 2} = \frac{z + 4}{z - 2} \] Thus, the results are: \[ fg(x) = \frac{1}{6x - 1} \quad \text{and} \quad gf(z) = \frac{z + 4}{z - 2} \]