a) \(f(x) = x\sqrt {4 - {x^2}} \), −2 ≤ x ≤ 2

Ta có: f'(x) = \(\sqrt {4 - {x^2}} + \frac{{ - {x^2}}}{{\sqrt {4 - {x^2}} }}\) = \(\frac{{4 - 2{x^2}}}{{\sqrt {4 - {x^2}} }}\);

           f'(x) = 0 ⇔ x = ±\(\sqrt 2 \).

Ta tính được các giá trị: f(−2) = f(2) = 0; f(−\(\sqrt 2 \)) = −2; f(\(\sqrt 2 \)) = 2.

Do đó, \(\mathop {\min }\limits_{\left[ { - 2;2} \right]} f\left( x \right) = f\left( { - \sqrt 2 } \right) = - 2\); \(\mathop {\max }\limits_{\left[ { - 2;2} \right]} f\left( x \right) = f\left( {\sqrt 2 } \right) = 2\).

b) f(x) = x – cosx, \( - \frac{\pi }{2} \le x \le \frac{\pi }{2}\)

Ta có: f'(x) = 1 + sinx

           f'(x) = 0 ⇔ 1 + sinx = 0 ⇔ x = \( - \frac{\pi }{2} + k2\pi \) (k ∈ ℤ).

Do \( - \frac{\pi }{2} \le x \le \frac{\pi }{2}\) nên x = \( - \frac{\pi }{2}\) (với k = 0).

Ta tính được các giá trị: \(f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2}\); \(f\left( { - \frac{\pi }{2}} \right) = \frac{\pi }{2}\).

Vậy \(\mathop {\min }\limits_{\left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right]} f\left( x \right) = f\left( { - \frac{\pi }{2}} \right) = - \frac{\pi }{2}\), \(\mathop {\max }\limits_{\left[ { - \frac{\pi }{2};\frac{\pi }{2}} \right]} f\left( x \right) = f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2}\).