Question

Question #333f1

Answer 1

There is a difference between inverse and inverse function.
See explanation for the solutions

Explanation:

`color(blue)("Case 1 - Inverse")`

`1/(h(x)) = 1/(-3x-3) -> -1/(3x+3)`

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`color(blue)("Case 2 - Inverse function")`

Set `h(x)=y=-3x-3`

Add 3 to both sides

`y+3=-3x`

Multiply both sides by (-1)

`-y-3=+3x`

Divide both sides by 3#

`-y/3-1=x`

Where there is `x` write `y` and where there is `y` write `x` giving:

`-x/3-1=y`

`y=-x/3-1`
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The inverse function is a reflection of the original function about `y=x`