Differentiate \( f(x) = \frac{1}{3} x^3 + 2 x^2 – 5 x \):

\[ f'(x) = x^2 + 4 x – 5 \]

Answer: \( f'(x) = x^2 + 4 x – 5 \).

To find the maximum, solve \( f'(x) = 0 \):

\[ x^2 + 4 x – 5 = 0 \]

Factorize:

\[ (x + 5)(x – 1) = 0 \]

\[ x = -5, \, x = 1 \]

To determine the maximum, consider the second derivative or graph behavior:

\[ f”(x) = 2 x + 4 \]

At \( x = -5 \), \( f”(-5) = 2(-5) + 4 = -6 < 0 \), indicating a maximum.

At \( x = 1 \), \( f”(1) = 2(1) + 4 = 6 > 0 \), indicating a minimum.

Thus, the maximum point M is at \( x = -5 \).

Answer: \( x = -5 \).

Method 1:

Find the second derivative:

\[ f”(x) = 2 x + 4 \]

Set \( f”(x) = 0 \) to find the point of inflection:

\[ 2 x + 4 = 0 \]

\[ x = -2 \]

Method 2:

Use symmetry of the cubic. The inflection point is at the midpoint of the maximum (\( x = -5 \)) and minimum (\( x = 1 \)):

\[ \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \]

Thus, the point of inflection N is at \( x = -2 \).

Answer: \( x = -2 \).

Find the slope of the tangent at \( x = 3 \):

\[ f'(x) = x^2 + 4 x – 5 \]

\[ f'(3) = 3^2 + 4 \cdot 3 – 5 = 9 + 12 – 5 = 16 \]

Use the point-slope form at \( (3, 12) \):

\[ y – 12 = 16 (x – 3) \]

\[ y – 12 = 16 x – 48 \]

\[ y = 16 x – 36 \]

Alternatively, use \( y = m x + b \):

\[ 12 = 16 \cdot 3 + b \]

\[ 12 = 48 + b \]

\[ b = -36 \]

\[ y = 16 x – 36 \]

Answer: \( y = 16 x – 36 \).