Respuesta :
Let's address each part of the question:
1. **Function \( h \):**
Given: \( h(x) = \frac{2}{3}x + 6 \)
A) To find \( h^{-1} \), first, we need to solve for \( x \) in terms of \( y \):
\[ y = \frac{2}{3}x + 6 \]
\[ y - 6 = \frac{2}{3}x \]
\[ x = \frac{3}{2}(y - 6) \]
So, the inverse function \( h^{-1}(y) \) is:
\[ h^{-1}(y) = \frac{3}{2}y - 9 \]
B)
i) To evaluate \( h^{-1}(12) \):
\[ h^{-1}(12) = \frac{3}{2} \times 12 - 9 = 18 - 9 = 9 \]
ii) To evaluate \( h^{-1}(-4) \):
\[ h^{-1}(-4) = \frac{3}{2} \times (-4) - 9 = -6 - 9 = -15 \]
2. **Function \( f \):**
Given: \( f(x) = \frac{x+1}{x-2} \)
A) The function \( f(x) \) is not defined when the denominator \( x - 2 \) equals zero. So, \( f(x) \) is not defined when \( x = 2 \).
B) To find \( f^{-1} \), we need to solve for \( x \) in terms of \( y \):
\[ y = \frac{x+1}{x-2} \]
\[ y(x-2) = x + 1 \]
\[ xy - 2y = x + 1 \]
\[ xy - x = 2y + 1 \]
\[ x(y - 1) = 2y + 1 \]
\[ x = \frac{2y + 1}{y - 1} \]
So, the inverse function \( f^{-1}(y) \) is:
\[ f^{-1}(y) = \frac{2y + 1}{y - 1} \]
ii) \( f^{-1} \) is not defined when \( y - 1 = 0 \), which means \( y = 1 \). Substituting \( y = 1 \) into \( f^{-1}(y) \) gives \( x = \frac{2(1) + 1}{1 - 1} = \frac{3}{0} \), which is undefined.
C) To evaluate \( f^{-1}(7) \):
\[ f^{-1}(7) = \frac{2(7) + 1}{7 - 1} = \frac{15}{6} = 2.5 \]
So, in summary:
- \( h^{-1}(y) = \frac{3}{2}y - 9 \)
- \( h^{-1}(12) = 9 \)
- \( h^{-1}(-4) = -15 \)
- \( f^{-1}(y) = \frac{2y + 1}{y - 1} \) (except when \( y = 1 \))
- \( f^{-1}(7) = 2.5 \)
