College Algebra Problems With Answers
Composite and Inverse Functions

• a) To find the inverse function \( f^{-1}(x) \):
  Replace \( f(x) \) with \( y \):
  \[ y = \sqrt{x - 4} + 3 \]
  Swap \( x \) and \( y \) to interchange the input and output roles:
  \[ x = \sqrt{y - 4} + 3 \]
  Isolate the square root term by subtracting 3 from both sides:
  \[ x - 3 = \sqrt{y - 4} \]
  Square both sides of the equation:
  \[ (x - 3)^2 = y - 4 \]
  Solve for \( y \) by adding 4 to both sides:
  \[ y = (x - 3)^2 + 4 \]
  Considering the range of the original function \( f(x) \), which is \([3, +\infty)\), the domain restriction for the inverse function is \( x \geq 3 \). Thus:
  \[ f^{-1}(x) = (x - 3)^2 + 4 \quad \text{for } x \geq 3 \]
• b) To find the range of \( f^{-1} \):
  The range of an inverse function is equal to the domain of the original function. Since the domain of \( f(x) = \sqrt{x - 4} + 3 \) requires \( x - 4 \geq 0 \) (giving \( x \geq 4 \)), the domain of \( f \) is \([4, +\infty)\). Therefore, the range of \( f^{-1} \) is: \[ [4, +\infty) \text{ } \]

• a) To find the inverse function \( h^{-1}(x) \):
  Write the function using \( y \):
  \[ y = \dfrac{x - 1}{-x + 3} \]
  Swap \( x \) and \( y \):
  \[ x = \dfrac{y - 1}{-y + 3} \]
  Multiply both sides by \((-y + 3)\) to clear the denominator:
  \[ x(-y + 3) = y - 1 \]
  Expand the left side:
  \[ -xy + 3x = y - 1 \]
  Rearrange terms to collect all terms containing \( y \) on one side:
  \[ 3x + 1 = y + xy \]
  Factor out \( y \) on the right side:
  \[ 3x + 1 = y(x + 1) \]
  Solve for \( y \):
  \[ y = \dfrac{3x + 1}{x + 1} \]
  Rewriting in standard fractional form gives:
  \[ h^{-1}(x) = \dfrac{-3x - 1}{x + 1} \text{ } \]
• b) To find the range of \( h \):
  The range of a function \( h \) is equal to the domain of its inverse \( h^{-1} \). Looking at \( h^{-1}(x) = \dfrac{-3x - 1}{x + 1} \), the denominator cannot be zero, meaning \( x + 1 \neq 0 \), or \( x \neq -1 \). Therefore, the domain of \( h^{-1} \) (and consequently the range of \( h \)) is: \[ (-\infty, -1) \cup (-1, +\infty) \text{ } \]

• a) To find \( (f \circ g)(x) \):
  By definition, \( (f \circ g)(x) = f(g(x)) \). Substitute \( g(x) = \dfrac{1}{x + 3} \) into \( f(x) \): \[ (f \circ g)(x) = \dfrac{ \dfrac{1}{x + 3} - 1}{ \dfrac{1}{x + 3} + 5} \] To simplify the complex fraction, multiply both the numerator and the denominator by \((x + 3)\): \[ (f \circ g)(x) = \dfrac{\left( \dfrac{1}{x + 3} - 1\right)(x + 3)}{\left( \dfrac{1}{x + 3} + 5\right)(x + 3)} = \dfrac{1 - (x + 3)}{1 + 5(x + 3)} \] Expand and simplify the numerator and denominator: \[ = \dfrac{1 - x - 3}{1 + 5x + 15} = \dfrac{-x - 2}{5x + 16} = - \dfrac{x + 2}{5x + 16} \text{ } \]
• b) To find the domain of \( f \circ g \):
  The domain of a composite function \( f \circ g \) includes all values of \( x \) in the domain of \( g \) such that \( g(x) \) is in the domain of \( f \):
  1. From the inner function \( g(x) = \dfrac{1}{x + 3} \), we must have \( x + 3 \neq 0 \), so \( x \neq -3 \).
  2. From the resulting expression \( \dfrac{-x - 2}{5x + 16} \), the denominator must be non-zero: \( 5x + 16 \neq 0 \implies x \neq - \dfrac{16}{5} \).
  Combining these restrictions, the domain of \( f \circ g \) is: \[ \left(-\infty, - \dfrac{16}{5}\right) \cup \left(- \dfrac{16}{5}, -3\right) \cup (-3, +\infty) \text{ } \]

To find the inverse function \( h^{-1}(x) \), follow these systematic steps:

Start with the equation defining \( h(x) \):
\[ y = f(x) + k \]
Swap \( x \) and \( y \) to find the inverse relation:
\[ x = f(y) + k \]
Isolate the function term \( f(y) \) by subtracting \( k \) from both sides:
\[ x - k = f(y) \]
Apply the inverse function \( f^{-1} \) to both sides of the equation:
\[ f^{-1}(x - k) = f^{-1}(f(y)) \]
Since \( f^{-1}(f(y)) = y \), we get:
\[ y = f^{-1}(x - k) \]
Therefore, the inverse function is:
\[ h^{-1}(x) = f^{-1}(x - k) \text{ } \]

To find the inverse function \( h^{-1}(x) \), follow these algebraic steps:

Start with the definition of \( h(x) \), replacing \( h(x) \) with \( y \):
\[ y = Af(x - h) + k \]
Swap \( x \) and \( y \) to begin finding the inverse:
\[ x = Af(y - h) + k \]
Isolate the term containing \( f(y - h) \) by subtracting \( k \) and dividing by \( A \):
\[ x - k = Af(y - h) \] \[ \dfrac{x - k}{A} = f(y - h) \]
Apply the inverse function \( f^{-1} \) to both sides to eliminate \( f \):
\[ f^{-1}\left( \dfrac{x - k}{A}\right) = f^{-1}(f(y - h)) \] \[ f^{-1}\left( \dfrac{x - k}{A}\right) = y - h \]
Solve for \( y \) by adding \( h \) to both sides:
\[ y = f^{-1}\left( \dfrac{x - k}{A}\right) + h \]
Therefore, the expression for the inverse function is:
\[ h^{-1}(x) = f^{-1}\left( \dfrac{x - k}{A}\right) + h \text{ } \]

• a) \( (f \circ g)(-4) = f(g(-4)) = f(2) = -2 \)
• b) \( (g \circ f)(1) = g(f(1)) = g(-3) = -1 \)

• a) \( (f \circ h)(1) = f(h(1)) = f(2) = 6 \)
• b) \( (f \circ f)(0) = f(f(0)) = f(1) = 2 \)
• c) \( (f \circ h)(5) = f(h(5)) = f(26) = \text{undefined} \)
• d) \( (f \circ h^{-1})(5) = f(h^{-1}(5)) = f(2) = 6 \)
• e) \( (h \circ f^{-1})(6) = h(f^{-1}(6)) = h(2) = 5 \)