Solution

(a) Intermediate Value Theorem Application:

\[ h(1) = f(g(1)) – 6 = f(2) – 6 = 9 – 6 = 3 \] \[ h(3) = f(g(3)) – 6 = f(4) – 6 = -1 – 6 = -7 \]

Since \( h \) is continuous (as \( f \) and \( g \) are differentiable) and \( -7 < -5 < 3 \), by the IVT there exists \( r \) in \( (1,3) \) such that \( h(r) = -5 \).

(b) Mean Value Theorem Application:

\[ \frac{h(3)-h(1)}{3-1} = \frac{-7-3}{2} = -5 \]

Since \( h \) is differentiable (by chain rule, as \( f \) and \( g \) are differentiable), by MVT there exists \( c \) in \( (1,3) \) where \( h'(c) = -5 \).

(c) Derivative Calculation:

Using the Fundamental Theorem of Calculus and chain rule: \[ w'(x) = f(g(x)) \cdot g'(x) \] \[ w'(3) = f(g(3)) \cdot g'(3) = f(4) \cdot 2 = -1 \cdot 2 = -2 \]

(d) Tangent Line to Inverse Function:

Since \( g(1) = 2 \), then \( g^{-1}(2) = 1 \).

The derivative of the inverse function: \[ (g^{-1})'(2) = \frac{1}{g'(g^{-1}(2))} = \frac{1}{g'(1)} = \frac{1}{5} \]

Tangent line equation: \[ y – 1 = \frac{1}{5}(x – 2) \]