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What is the inverse of the function {:[h(x)=(5+x)/(4-2x)?],[h^(-1)(x)=]:}

What is the inverse of the function\newlineh(x)=5+x4−2x?h−1(x)=□ \begin{array}{l} h(x)=\frac{5+x}{4-2 x} ? \\ h^{-1}(x)=\square \end{array}

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Full solution

Q. What is the inverse of the function\newlineh(x)=5+x4−2x?h−1(x)=□ \begin{array}{l} h(x)=\frac{5+x}{4-2 x} ? \\ h^{-1}(x)=\square \end{array}
  1. Switching roles and setting up the equation: To find the inverse of the function h(x)=5+x4−2xh(x) = \frac{5+x}{4-2x}, we need to switch the roles of xx and h(x)h(x) and then solve for the new xx.
    Let y=5+x4−2xy = \frac{5+x}{4-2x}.
    Now, switch xx and yy to get x=5+y4−2yx = \frac{5+y}{4-2y}.
  2. Multiplying both sides to eliminate the fraction: Next, we need to solve for yy in terms of xx. To do this, we'll multiply both sides of the equation by (4−2y)(4-2y) to get rid of the fraction.\newline(4−2y)x=5+y(4-2y)x = 5+y
  3. Distributing xx on the left side: Now, distribute xx on the left side of the equation.4x−2xy=5+y4x - 2xy = 5 + y
  4. Moving terms to separate sides: Next, we want to get all terms involving yy on one side and the constant terms on the other side. Let's move −2xy-2xy to the right side and 55 to the left side.\newline4x−5=y+2xy4x - 5 = y + 2xy
  5. Factoring out yy: Now, factor out yy on the right side of the equation.4x−5=y(1+2x)4x - 5 = y(1 + 2x)
  6. Isolating y: To isolate y, divide both sides of the equation by (1+2x)(1 + 2x).\newliney=4x−51+2xy = \frac{4x - 5}{1 + 2x}
  7. Solving for the inverse function: We have now solved for yy in terms of xx, which gives us the inverse function.h−1(x)=4x−51+2xh^{-1}(x) = \frac{4x - 5}{1 + 2x}

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