QQuestionMathematics
QuestionMathematics
The one to one function h is defined below. h(2$)= 7x- 2 /x+ 1 Find h^- 1(x), where h^- 1 is the inverse of h. Also state the domain and range of h^- 1 in interval notation.
12 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1:: Set the original function h(x) equal to y.
y = h(x) = \frac{7x-2}{x+1}
Step 2:: To find the inverse function, switch x and y.
x = h^{-1}(y) = \frac{7y-2}{y+1}
Step 3:: Solve for y to get h^{- 1}(y) in terms of x.
y = \frac{xy + 2}{7}
Step 4:: Replace h^{- 1}(y) with h^{- 1}(x) since we are looking for the inverse function.
h^{-1}(x) = \frac{xh^{-1}(x) + 2}{7}
Step 5:: Solve for h^{- 1}(x).
h^{-1}(x) = \frac{2}{7-x}
Step 6:: Find the domain of h^{- 1}(x).
(-\infty, 7) \cup (7, \infty)
The domain of a function is the set of all possible inputs. For h^{- 1}(x), we need to find the values of x that make the denominator nonzero. So, we need to find when 7 -x is not equal to 0. So, the domain of h^{- 1}(x) is all real numbers except x = 7. In interval notation, this is written as:
Step 7:: Find the range of h^{- 1}(x).
(-\infty, \infty)
The range of a function is the set of all possible outputs. Since h(2$) has a horizontal asymptote at y = 7, h^{- 1}(x) will have a vertical asymptote at x = 7. This means that as x approaches 7, h^{- 1}(x) will approach positive or negative infinity. Therefore, the range of h^{- 1}(x) is all real numbers. In interval notation, this is written as:
Final Answer
The inverse function is: h^{- 1}(x) = \frac{2}{7 -x} The domain of h^{- 1}(x) is: (-\infty, 7) \cup (7, \infty) The range of h^{- 1}(x) is: (-\infty, \infty)
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students