kbac98
Answered

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Find the two x-intercepts of the function f and show that f '(x) = 0 at some point between the two x-intercepts.
[tex]f(x)=x\sqrt{x+8}[/tex]

Sagot :

jasongggg

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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