If the function f (x)=x(x+3) e ^ - x 2 satisfies all the conditions of Rolle's theorem in [-3,0] , then c is

Step-by-step solution

Rolle's Theorem requires finding c (-3,0) such that f'(c) = 0 . The function f(x) = x(x+3)e^ -x/2 satisfies the theorem's conditions: it is continuous on [-3,0] , differentiable on (-3,0) , and f(-3) = f(0) = 0 . The derivative is computed using the product rule: f'(x) = d dx [x(x+3) ]e^ -x/2 + x(x+3) d dx [e^ -x/2 ] f'(x) = (2x+3)e^ -x/2 - 1 2 x(x+3)e^ -x/2 Factoring yields: f'(x) = e^ -x/2 (- 1 2 x^2 + 1 2 x + 3 ) Setting f'(c) = 0 and noting e^ -c/2 0 : - 1 2 c^2 + 1 2 c + 3 = 0 Multiplying through by -2 gives c