Find the two x-intercepts of the function f and show that f '(x) = 0 at some point between the two x-intercepts.
1 answer:
Step-by-step explanation:
f(x) = x√(x+8)
When f(x) = 0,
0 = x√(x+8)
x = 0, x + 8 = 0
x = -8
f'(x) = (1)√(x+8) + x • ½(x+8)^(-½) = 0
0 = √(x+8) + 1/[2√(x+8)] (x)
-2(x+8) = x
-2x - 16 = x
x = -16/3
(-8 < -16/3 < 0)
Therefore, x-intercept of f'(x) = 0 is somewhere between the two x-intercepts, ranging from -8 to 0.
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