Differentiate \( f(x) = ax^3 + bx^2 + cx + d \).

\( \frac{d}{dx}(ax^3) = 3ax^2 \), \( \frac{d}{dx}(bx^2) = 2bx \), \( \frac{d}{dx}(cx) = c \), \( \frac{d}{dx}(d) = 0 \).

\[ f'(x) = 3ax^2 + 2bx + c \]

Answer: \( f'(x) = 3ax^2 + 2bx + c \).

Show \( b^2 – 3ac > 0 \) given \( f^{-1} \) does not exist.

For \( f^{-1} \) to not exist, \( f(x) \) is not one-to-one, so \( f'(x) = 3ax^2 + 2bx + c = 0 \) must have two real roots (indicating two turning points).

Discriminant of the quadratic: \( \Delta = (2b)^2 – 4 \cdot 3a \cdot c = 4b^2 – 12ac \).

For two real roots: \( 4b^2 – 12ac > 0 \).

\[ b^2 – 3ac > 0 \]

Answer: \( b^2 – 3ac > 0 \), as shown.

Show \( g^{-1} \) exists for \( g(x) = \frac{1}{2}x^3 – 3x^2 + 6x – 8 \).

Compute \( g'(x) \): \( \frac{d}{dx} \left( \frac{1}{2}x^3 \right) = \frac{3}{2}x^2 \), \( \frac{d}{dx}(-3x^2) = -6x \), \( \frac{d}{dx}(6x) = 6 \), \( \frac{d}{dx}(-8) = 0 \).

\[ g'(x) = \frac{3}{2}x^2 – 6x + 6 \]

Discriminant: \( \Delta = (-6)^2 – 4 \cdot \frac{3}{2} \cdot 6 = 36 – 36 = 0 \).

One real root implies one critical point. Since \( g(x) \) is cubic, this is an inflection point, and \( g(x) \) is strictly increasing (monotonic), so \( g^{-1} \) exists.

Answer: \( g^{-1} \) exists, as shown.

Write \( g(x) = \frac{1}{2}x^3 – 3x^2 + 6x – 8 \) as \( p (x – 2)^3 + q \).

Expand: \( (x – 2)^3 = x^3 – 6x^2 + 12x – 8 \).

\[ \frac{1}{2}(x – 2)^3 = \frac{1}{2}x^3 – 3x^2 + 6x – 4 \]

\[ g(x) = \frac{1}{2}(x – 2)^3 – 4 \]

Thus, \( p = \frac{1}{2} \), \( q = -4 \).

Answer: \( p = \frac{1}{2} \), \( q = -4 \).

Find \( g^{-1}(x) \).

From (b)(ii): \( g(x) = \frac{1}{2}(x – 2)^3 – 4 \).

Set \( y = g(x) \):
\[ y = \frac{1}{2}(x – 2)^3 – 4 \]

Solve for \( x \):
\[ y + 4 = \frac{1}{2}(x – 2)^3 \]
\[ (x – 2)^3 = 2(y + 4) \]
\[ x – 2 = \sqrt[3]{2(y + 4)} \]
\[ x = \sqrt[3]{2(y + 4)} + 2 \]

Thus, \( g^{-1}(x) = \sqrt[3]{2(x + 4)} + 2 \).

Answer: \( g^{-1}(x) = \sqrt[3]{2(x + 4)} + 2 \).

State the three transformations from \( y = x^3 \) to \( y = g(x) \).

\( g(x) = \frac{1}{2}(x – 2)^3 – 4 \).

1. Translate by \( \begin{pmatrix} 2 \\ 0 \end{pmatrix} \): \( y = x^3 \to y = (x – 2)^3 \).

2. Scale by \( \frac{1}{2} \) parallel to the y-axis: \( y = (x – 2)^3 \to y = \frac{1}{2}(x – 2)^3 \).

3. Translate by \( \begin{pmatrix} 0 \\ -4 \end{pmatrix} \): \( y = \frac{1}{2}(x – 2)^3 \to y = \frac{1}{2}(x – 2)^3 – 4 \).

Answer: Translation by \( \begin{pmatrix} 2 \\ 0 \end{pmatrix} \), stretch by scale factor \( \frac{1}{2} \) parallel to the y-axis, translation by \( \begin{pmatrix} 0 \\ -4 \end{pmatrix} \).

Sketch \( y = g(x) \) and \( y = g^{-1}(x) \), indicating axis intercepts.

For \( g(x) \): x-intercepts via \( \frac{1}{2}x^3 – 3x^2 + 6x – 8 = 0 \). Test roots (e.g., \( x = 2 \)): \( \frac{1}{2} \cdot 8 – 3 \cdot 4 + 6 \cdot 2 – 8 = 4 – 12 + 12 – 8 = -4 \). Numerical or graphical methods suggest intercepts around \( x \approx 4.64 \). y-intercept: \( g(0) = -8 \).

For \( g^{-1}(x) \): x-intercept at \( g^{-1}(0) = \sqrt[3]{2 \cdot 4} + 2 = \sqrt[3]{8} + 2 = 4 \). y-intercept via \( g(y) = 0 \), so \( y \approx 4.64 \).

Answer: Graphs with intercepts: \( g(x) \) at \( (4.64, 0) \), \( (0, -8) \); \( g^{-1}(x) \) at \( (0, 4.64) \), \( (4, 0) \).