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

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

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Full solution

Q. What is the inverse of the function\newlineh(x)=−2x−1x+5?h−1(x)= \begin{array}{l} h(x)=\frac{-2 x-1}{x+5} ? \\ h^{-1}(x)= \end{array}
  1. Rewrite function as 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 rewriting the function h(x)h(x) as yy:
    y=−2x−1x+5y = \frac{-2x - 1}{x + 5}
    Now, we replace yy with xx and xx with yy to find the inverse function:
    h(x)h(x)22
  2. Replace xx and yy: Next, we need to solve for yy. To do this, we'll multiply both sides of the equation by (y+5)(y + 5) to eliminate the denominator: x(y+5)=−2y−1x(y + 5) = -2y - 1
  3. Eliminate denominator: Now, distribute xx on the left side of the equation: xy+5x=−2y−1xy + 5x = -2y - 1
  4. Distribute xx: To isolate yy, we need to get all the terms with yy on one side of the equation and the constant terms on the other side. Let's move the terms with yy to the left side and the constant terms to the right side:\newlinexy+2y=−5x−1xy + 2y = -5x - 1
  5. Isolate yy: Now, factor out yy on the left side of the equation:\newliney(x+2)=−5x−1y(x + 2) = -5x - 1
  6. Factor out yy: Finally, divide both sides of the equation by (x+2)(x + 2) to solve for yy:y=−5x−1x+2y = \frac{-5x - 1}{x + 2}This is the inverse function of h(x)h(x).

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