Find the two x-intercepts of the function f and show that f '(x) = 0 at some point between the two x-intercepts. f(x) = −8x sqrt(x + 3)

Question

Find the two x-intercepts of the function f and show that f '(x) = 0 at some point between the two x-intercepts.
f(x) = −8x sqrt(x + 3)

My two intercepts thus far are (0,0) and (-3,0)

But I need help with finding a value of x such that f '(x) = 0.

Answer 1

f(x) is continuous on [-3,0] and is differentiable on (-3,0)

 

f(-3) = f(0) = 0

 

By Rolle's Theorem, there exists at least one value, c, in the interval (-3,0) such that f'(c) = 0.

 

f'(x) = -8√(x+3) - 8x(1/2)(x+3)^-1/2

 

       = -8√(x+3) -4x / √(x+3)

 

       = [-8(x+3) - 4x] / √(x+3) = [-12x-24] / √(x+3)

 

f'(c) = 0 when -12c - 24 = 0

 

So, c = -2

 

Answer 2

Use the Zero Product Rule (Algebra 2)

f(x) = −8x √(x + 3)

0 = −8x √(x + 3)

0 = -8x, x = 0

or

0 = √(x + 3), x = -3