Transcription 15/11/2025, 10:41:02 PM

Notes

• Function Definition: Inverse of h(x) Let h(x) = \(\frac{6x + 5}{7x - 3}\). a. Fill in the blanks with the correct numbers. To find \(h^{-1}(x)\), we start by setting \(h(x) = y\): \[ y = \frac{6x + 5}{7x - 3} \] Now, solve for x in terms of y: \[ y(7x - 3) = 6x + 5 \] \[ 7xy - 3y = 6x + 5 \] Bring all x terms to one side: \[ 7xy - 6x = 3y + 5 \] Factor x out: \[ x(7y - 6) = 3y + 5 \] Solve for x: \[ x = \frac{3y + 5}{7y - 6} \] Since the inverse function swaps x and y, the inverse function is: \[ h^{-1}(x) = \frac{3x + 5}{7x - 6} \] **Answer:** \[ h^{-1}(x) = \frac{\boxed{3x + 5}}{\boxed{7x - 6}} \] --- • Composition of Functions: \((h \circ h^{-1})(x)\) b. Which of the following is equal to \((h \circ h^{-1})(x)\)? Options: \(-x\), \(x - 3\), \(x\), \(x + 5\), \(x^2\) Since the composition of a function and its inverse always equals the original input (for all x in the domain), the answer is: **Answer:** \[ \boxed{x} \] --- **Summary:** - The inverse function is \(h^{-1}(x) = \frac{3x + 5}{7x - 6}\). - The composition \((h \circ h^{-1})(x) = x\).