Question

score: 26.86/33 answered: 9/12
question 10
given the function $h(x) = \frac{9x - 13}{10x - 11}$, determine the inverse function $h^{-1}(x)$ in simplified form:
$h^{-1}(x) = \square$
determine the domain and range for both $h(x)$ and $h^{-1}(x)$ using interval notation:
domain of $h(x)$: $\square$
domain of $h^{-1}(x)$: $\square$
range of $h(x)$: $\square$
range of $h^{-1}(x)$: $\square$
question help: video

Explanation:

Step1: Find the inverse function

Let \( y = h(x) = \frac{9x - 13}{10x - 11} \). Swap \( x \) and \( y \): \( x = \frac{9y - 13}{10y - 11} \).
Multiply both sides by \( 10y - 11 \): \( x(10y - 11) = 9y - 13 \).
Expand: \( 10xy - 11x = 9y - 13 \).
Bring terms with \( y \) to one side: \( 10xy - 9y = 11x - 13 \).
Factor \( y \): \( y(10x - 9) = 11x - 13 \).
Solve for \( y \): \( y = \frac{11x - 13}{10x - 9} \), so \( h^{-1}(x) = \frac{11x - 13}{10x - 9} \).

Step2: Domain of \( h(x) \)

For \( h(x) = \frac{9x - 13}{10x - 11} \), denominator cannot be zero: \( 10x - 11
eq 0 \Rightarrow x
eq \frac{11}{10} \). So domain is \( (-\infty, \frac{11}{10}) \cup (\frac{11}{10}, \infty) \).

Step3: Domain of \( h^{-1}(x) \)

For \( h^{-1}(x) = \frac{11x - 13}{10x - 9} \), denominator \( 10x - 9
eq 0 \Rightarrow x
eq \frac{9}{10} \). Domain: \( (-\infty, \frac{9}{10}) \cup (\frac{9}{10}, \infty) \).

Step4: Range of \( h(x) \)

The range of \( h(x) \) is the domain of \( h^{-1}(x) \), so \( (-\infty, \frac{9}{10}) \cup (\frac{9}{10}, \infty) \).

Step5: Range of \( h^{-1}(x) \)

The range of \( h^{-1}(x) \) is the domain of \( h(x) \), so \( (-\infty, \frac{11}{10}) \cup (\frac{11}{10}, \infty) \).

Answer:

\( h^{-1}(x) = \boldsymbol{\frac{11x - 13}{10x - 9}} \)
Domain of \( h(x) \): \( \boldsymbol{(-\infty, \frac{11}{10}) \cup (\frac{11}{10}, \infty)} \)
Domain of \( h^{-1}(x) \): \( \boldsymbol{(-\infty, \frac{9}{10}) \cup (\frac{9}{10}, \infty)} \)
Range of \( h(x) \): \( \boldsymbol{(-\infty, \frac{9}{10}) \cup (\frac{9}{10}, \infty)} \)
Range of \( h^{-1}(x) \): \( \boldsymbol{(-\infty, \frac{11}{10}) \cup (\frac{11}{10}, \infty)} \)