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What is the inverse of the function {:[h(x)=(5x-3)/(x-1)?],[h^(-1)(x)=]:}

What is the inverse of the function\newlineh(x)=5x−3x−1?h−1(x)=□ \begin{array}{l} h(x)=\frac{5 x-3}{x-1} ? \\ h^{-1}(x)=\square \end{array}

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Full solution

Q. What is the inverse of the function\newlineh(x)=5x−3x−1?h−1(x)=□ \begin{array}{l} h(x)=\frac{5 x-3}{x-1} ? \\ h^{-1}(x)=\square \end{array}
  1. Replace with y: To find the inverse of the function h(x)h(x), we need to switch the roles of xx and yy in the equation and then solve for yy. Let's start by replacing h(x)h(x) with yy:y=5x−3x−1y = \frac{5x - 3}{x - 1}
  2. Switch x and y: Now we switch x and y to find the inverse: x=5y−3y−1x = \frac{5y - 3}{y - 1}
  3. Multiply and solve: Next, we solve for yy. Multiply both sides by (y−1)(y - 1) to get rid of the fraction:\newlinex(y−1)=5y−3x(y - 1) = 5y - 3
  4. Distribute xx: Distribute xx on the left side of the equation:\newlinexy−x=5y−3xy - x = 5y - 3
  5. Move 5y5y term: Now, we want to get all the terms with yy on one side and the constants on the other. Let's move the 5y5y term to the left side by subtracting 5y5y from both sides: xy−5y=x+3xy - 5y = x + 3
  6. Factor out yy: Factor out yy on the left side of the equation:\newliney(x−5)=x+3y(x - 5) = x + 3
  7. Divide and solve: Finally, divide both sides by (x−5)(x - 5) to solve for yy:y=x+3x−5y = \frac{x + 3}{x - 5}
  8. Final Inverse Function: We have found the inverse function of h(x)h(x). The inverse function is:\newlineh−1(x)=x+3x−5h^{-1}(x) = \frac{x + 3}{x - 5}

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