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

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

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Q. What is the inverse of the function\newlineh(x)=3−xx+1 ? h−1(x)= \begin{array}{l} h(x)=\frac{3-x}{x+1} \text { ? } \\ h^{-1}(x)= \end{array}
  1. Replace h(x)h(x) with yy: 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=3−xx+1y = \frac{3 - x}{x + 1}
  2. Switch x and y: Now we switch x and y to find the inverse: x=3−yy+1x = \frac{3 - y}{y + 1}
  3. Multiply both sides by (y+1)(y + 1): Next, we solve for yy. Multiply both sides by (y+1)(y + 1) to get rid of the fraction:\newlinex(y+1)=3−yx(y + 1) = 3 - y\newlinexy+x=3−yxy + x = 3 - y
  4. Move the term with \newliney to the left side: Now, we need to get all the terms with \newliney on one side and the constant terms on the other side. Let's move the term with \newliney from the right side to the left side:\newline\newlinexy + y = 33 - x
  5. Factor out yy from the left side: Factor out yy from the left side:\newliney(x+1)=3−xy(x + 1) = 3 - x
  6. Divide both sides by (x+1)(x + 1): Now, divide both sides by (x+1)(x + 1) to solve for yy:y=3−xx+1y = \frac{3 - x}{x + 1}
  7. Error in solving for yy: The correct division to isolate yy is:\newliney=3−xx+1y = \frac{3 - x}{x + 1}\newlineHowever, this is the original function, not its inverse. We made an error in the process of solving for yy. We need to go back and correctly isolate yy.

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