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Question Show that the inverse function f−1(x)=8x−83f^{-1}(x) = \frac{8 x-8}{3} of f(x)=38x+1f(x) = \frac{3}{8} x+1 using function composition.

Studdy Solution

STEP 1

Assumptions1. The function ff is given by f(x)=38x+1f(x)=\frac{3}{8} x+1 . The inverse function f−1f^{-1} is given by f−1(x)=8x−83f^{-1}(x)=\frac{8 x-8}{3}
3. The composition of a function and its inverse should yield the identity function, i.e., f(f−1(x))=xf(f^{-1}(x))=x and f−1(f(x))=xf^{-1}(f(x))=x

STEP 2

First, we will verify that f(f−1(x))=xf(f^{-1}(x))=x. We will substitute f−1(x)f^{-1}(x) into f(x)f(x).
f(f−1(x))=f(8x−8)f(f^{-1}(x)) = f\left(\frac{8 x-8}{}\right)

STEP 3

Now, we will replace xx in f(x)f(x) with 8x−83\frac{8 x-8}{3}.
f(f−1(x))=38⋅8x−83+1f(f^{-1}(x)) = \frac{3}{8} \cdot \frac{8 x-8}{3} +1

STEP 4

implify the expression by cancelling out the3 in the numerator and the denominator, and the8 in the numerator and the denominator.
f(f−1(x))=x−1+1f(f^{-1}(x)) = x -1 +1

STEP 5

implify the expression further.
f(f−1(x))=xf(f^{-1}(x)) = x

STEP 6

Next, we will verify that f−1(f(x))=xf^{-1}(f(x))=x. We will substitute f(x)f(x) into f−1(x)f^{-1}(x).
f−1(f(x))=f−1(38x+1)f^{-1}(f(x)) = f^{-1}\left(\frac{3}{8} x+1\right)

STEP 7

Now, we will replace xx in f−1(x)f^{-1}(x) with 3x+1\frac{3}{} x+1.
f−1(f(x))=⋅3x+3f^{-1}(f(x)) = \frac{ \cdot \frac{3}{} x+}{3}

STEP 8

implify the expression by cancelling out the8 in the numerator and the denominator.
f−1(f(x))=3x+83f^{-1}(f(x)) = \frac{3x+8}{3}

STEP 9

implify the expression further by dividing each term in the numerator by3.
f−(f(x))=x+83−83f^{-}(f(x)) = x + \frac{8}{3} - \frac{8}{3}

STEP 10

implify the expression further.
f−(f(x))=xf^{-}(f(x)) = xSince both f(f−(x))=xf(f^{-}(x))=x and f−(f(x))=xf^{-}(f(x))=x are true, we have shown that f−(x)=8x−83f^{-}(x)=\frac{8 x-8}{3} is indeed the inverse of f(x)=38x+f(x)=\frac{3}{8} x+.

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