a) To find \( (f \circ g)(x) \):
Given: \( f(x) = 2x – x^3 \), \( g(x) = \tan x \).
Composition: \( (f \circ g)(x) = f(g(x)) \).
Substitute: \( f(\tan x) = 2 \times \tan x – (\tan x)^3 \).
Simplify:
\( (f \circ g)(x) = 2 \tan x – \tan^3 x \) [2]

b) To sketch the graph of \( y = (f \circ g)(x) = 2 \tan x – \tan^3 x \) for \( -1 \leq x \leq 1 \) and find local maxima and minima:

Properties: Odd function, passes through \( (0, 0) \), since \( \tan 0 = 0 \), so \( y = 2 \times 0 – 0^3 = 0 \).
Domain: \( -1 \leq x \leq 1 \), asymptotes at \( x = \pm \frac{\pi}{2} \approx \pm 1.571 \) are outside.

Find extrema by differentiating:
\( y = 2 \tan x – \tan^3 x \)
Derivative: \( y’ = 2 \times \sec^2 x – 3 \times \tan^2 x \times \sec^2 x \).
Factor: \( y’ = \sec^2 x \times (2 – 3 \tan^2 x) \).
Set \( y’ = 0 \):
\( 2 – 3 \tan^2 x = 0 \)
\( \tan^2 x = \frac{2}{3} \)
\( \tan x = \pm \sqrt{\frac{2}{3}} \approx \pm 0.8165 \)
\( x \approx \pm 0.685 \) radians

Evaluate y at critical points:
For \( x \approx 0.685 \), \( \tan x \approx 0.8165 \):
\( y \approx 2 \times 0.8165 – (0.8165)^3 \approx 1.633 – 0.544 \approx 1.09 \)
For \( x \approx -0.685 \), \( \tan x \approx -0.8165 \):
\( y \approx 2 \times (-0.8165) – (-0.8165)^3 \approx -1.633 + 0.544 \approx -1.09 \)

Confirm extrema with second derivative test:
\( y’ = \sec^2 x \times (2 – 3 \tan^2 x) \)
Second derivative is complex, but sign change of \( y’ \) around critical points confirms:
Local maximum at \( (0.685, 1.09) \), local minimum at \( (-0.685, -1.09) \).

Graph features:
– Passes through \( (0, 0) \).
– Increases to maximum at \( (0.685, 1.09) \).
– Decreases to minimum at \( (-0.685, -1.09) \).
– Endpoints at \( x = \pm 1 \), where \( \tan 1 \approx 1.557 \), so \( y \approx 2 \times 1.557 – (1.557)^3 \approx -0.663 \).

Coordinates of extrema:
Local maximum: \( (0.685, 1.09) \)
Local minimum: \( (-0.685, -1.09) \) [3]