Let g(x)= x squareroot(4-x^2)

Question

Let g(x)= x squareroot(4-x^2)

a) Find g'(x)

b) Find the critical numbers of g.

c) Find the absolute maximum and absolute minimum values of g on the interval [-2,6/5]

Answer 1

f'(x) = √(4-x²) - x²/√(4-x²) = (4 - 2x²)/√(4-x²) -

4 - 2x² = 0 when x = ±√2

4-x² = 0 when x = ±2;

-2 is one of the endpoints. As √2 and 2 > 6/5, these are not critical points.

Thus, our critical point is -√2 and our endpoints are -2 and 6/5. You have to go with your book's definition in regard to -2 and 6/5.

f(x) = x√(4-x²), so

f(-2) = 0.

f(-√2) = -√2√(4-(-√2)²) = -√2√(4-2) = -2

f(6/5) = 6/5√(4-(6/5)²) = 6/5 * 8/5 = 48/25

The minimum is -2 at -√2. The maximum is 48/25 at 6/5