Questions: The function f(x)=(2x+1)/(x-5) is one-to-one. a. Find an equation for f^(-1)(x), the inverse function. b. Verify that your equation is correct by showing that f(f^(-1)(x))=x and f^(-1)(f(x))=, a. Select the correct choice below and fill in the answer box(es) to complete your choice (Simplify your answers. Use integers or fractions for any numbers in the expression.) A. f^(-1)(x)= , for x ≠ B. f^(-1)(x)= , for all x C. f^(-1)(x)= , for x ≤ D. f^(-1)(x)= , for x ≥

$The function f(x)=(2x+1)/(x-5) is one-to-one. a. Find an equation for f^(-1)(x), the inverse function. b. Verify that your equation is correct by showing that f(f^(-1)(x))=x and f^(-1)(f(x))=, a. Select the correct choice below and fill in the answer box(es) to complete your choice (Simplify your answers. Use integers or fractions for any numbers in the expression.) A. f^(-1)(x)= , for x ≠ B. f^(-1)(x)= , for all x C. f^(-1)(x)= , for x ≤ D. f^(-1)(x)= , for x ≥$

Transcript text: The function $f(x)=\frac{2 x+1}{x-5}$ is one-to-one. a. Find an equation for $f^{-1}(x)$, the inverse function. b. Verify that your equation is correct by showing that $f\left(f^{-1}(x)\right)=x$ and $f^{-1}(f(x))=$, a. Select the correct choice below and fill in the answer box(es) to complete your che (Simplify your answers. Use integers or fractions for any numbers in the expression.) A. $f^{-1}(x)=$ $\square$ , for $\mathrm{x} \neq$ $\square$ B. $f^{-1}(x)=$ $\square$ , for all $x$ C. $f^{-1}(x)=$ $\square$ , for $\mathrm{x} \leq$ $\square$ D. $f^{-1}(x)=$ $\square$ , for $x \geq$ $\square$

Solution

Solution Steps

Solution Approach

To find the inverse of the function $ f(x) = \frac{2x+1}{x-5} $, we need to solve the equation $ y = \frac{2x+1}{x-5} $ for $ x $ in terms of $ y $. This involves swapping $ x $ and $ y $ and then solving for the new $ y $. After finding the inverse function, we verify it by checking if $ f(f^{-1}(x)) = x $ and $ f^{-1}(f(x)) = x $.

Step 1: Finding the Inverse Function

To find the inverse of the function $ f(x) = \frac{2x + 1}{x - 5} $, we start by setting $ y = f(x) $: \[ y = \frac{2x + 1}{x - 5} \] Next, we swap $ x $ and $ y $ to solve for $ y $: \[ x = \frac{2y + 1}{y - 5} \] Multiplying both sides by $ (y - 5) $ gives: \[ x(y - 5) = 2y + 1 \] Expanding and rearranging leads to: \[ xy - 5x = 2y + 1 \implies xy - 2y = 5x + 1 \implies y(x - 2) = 5x + 1 \] Finally, solving for $ y $ yields: \[ y = \frac{5x + 1}{x - 2} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{5x + 1}{x - 2} \]

Step 2: Verifying the Inverse Function

To verify that $ f(f^{-1}(x)) = x $, we substitute $ f^{-1}(x) $ into $ f(x) $: \[ f(f^{-1}(x)) = f\left(\frac{5x + 1}{x - 2}\right) = \frac{2\left(\frac{5x + 1}{x - 2}\right) + 1}{\left(\frac{5x + 1}{x - 2}\right) - 5} \] Calculating the numerator: \[ 2\left(\frac{5x + 1}{x - 2}\right) + 1 = \frac{10x + 2}{x - 2} + \frac{x - 2}{x - 2} = \frac{10x + 2 + x - 2}{x - 2} = \frac{11x}{x - 2} \] Calculating the denominator: \[ \left(\frac{5x + 1}{x - 2}\right) - 5 = \frac{5x + 1 - 5(x - 2)}{x - 2} = \frac{5x + 1 - 5x + 10}{x - 2} = \frac{11}{x - 2} \] Thus, we have: \[ f(f^{-1}(x)) = \frac{\frac{11x}{x - 2}}{\frac{11}{x - 2}} = x \] This confirms that $ f(f^{-1}(x)) = x $.

Step 3: Verifying the Other Direction

Now we check $ f^{-1}(f(x)) $: \[ f^{-1}(f(x)) = f^{-1}\left(\frac{2x + 1}{x - 5}\right) = \frac{5\left(\frac{2x + 1}{x - 5}\right) + 1}{\left(\frac{2x + 1}{x - 5}\right) - 2} \] Calculating the numerator: \[ 5\left(\frac{2x + 1}{x - 5}\right) + 1 = \frac{10x + 5}{x - 5} + \frac{x - 5}{x - 5} = \frac{10x + 5 + x - 5}{x - 5} = \frac{11x}{x - 5} \] Calculating the denominator: \[ \left(\frac{2x + 1}{x - 5}\right) - 2 = \frac{2x + 1 - 2(x - 5)}{x - 5} = \frac{2x + 1 - 2x + 10}{x - 5} = \frac{11}{x - 5} \] Thus, we have: \[ f^{-1}(f(x)) = \frac{\frac{11x}{x - 5}}{\frac{11}{x - 5}} = x \] This confirms that $ f^{-1}(f(x)) = x $.