Question

consider the function f(x) = 3/(x - 1) for x > 1. (a) find f^(-1)(2) (b) use theorem 7, page 156 of the stewart essential calculus textbook to find (f^(-1))(2) (c) calculate f^(-1)(x) and state domain and range of f^(-1). use interval notation. if needed enter inf for ∞ or -inf for -∞. domain = range = calculate (f^(-1))(2) from the formula for f^(-1)(x) and check that it agrees with the result of part (b)

Explanation:

Step1: Find the inverse of the function

Let $y = f(x)=\frac{3}{x - 1},x>1$. Swap $x$ and $y$: $x=\frac{3}{y - 1}$. Then solve for $y$: $x(y - 1)=3$, $xy-x = 3$, $xy=x + 3$, $y=f^{-1}(x)=\frac{x + 3}{x}=1+\frac{3}{x},x>0$.

Step2: Find $f^{-1}(2)$

Substitute $x = 2$ into $f^{-1}(x)$: $f^{-1}(2)=1+\frac{3}{2}=\frac{2 + 3}{2}=\frac{5}{2}$.

Step3: Differentiate the inverse - function

First, rewrite $f^{-1}(x)=1 + 3x^{-1}$. Then, using the power rule $(x^n)'=nx^{n - 1}$, we have $(f^{-1})'(x)=-3x^{-2}=-\frac{3}{x^{2}}$.

Step4: Calculate $(f^{-1})'(2)$

Substitute $x = 2$ into $(f^{-1})'(x)$: $(f^{-1})'(2)=-\frac{3}{4}$.

Answer:

$(f^{-1})(2)=\frac{5}{2}$, $(f^{-1})'(2)=-\frac{3}{4}$, $f^{-1}(x)=1+\frac{3}{x}$, Domain of $f^{-1}=(0,\infty)$, Range of $f^{-1}=(1,\infty)$