SOLUTION: Find the critical points of the following function. f(x)=x^2sqrt(x+14)

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Question 1155423: Find the critical points of the following function.
f(x)=x^2sqrt(x+14)

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose that x=c is a critical point of f+%28x+%29.
If f'%28x+%29%3E0 to the left of x=c and f'%28x+%29%3C0 to the right of x=c, then x=c is a local maximum.
If f'%28x+%29%3C0 to the left of x=c and f'%28x+%29%3E0+to the right of +x=c, then x=c is a local minimum.
If f'%28x+%29 is the same sign on both sides of x=c then x=c is neither a local maximum nor a local minimum.

f%28x%29=x%5E2%2Asqrt%28x%2B14%29.....derivate
apply the Product Rule: %28f%2A+g+%29'=f '* g+f*+g'

f%C2%A2%28x%29=2x%2Asqrt%28x%2B14%29%2Bx%5E2%2A1%2F%282sqrt%28x%2B14%29%29

f%C2%A2%28x%29=%284x%2A%28x%2B14%29%2Bx%5E2%29%2F%282sqrt%28x%2B14%29%29
f%C2%A2%28x%29=%284x%5E2%2B56%2Bx%5E2%29%2F%282sqrt%28x%2B14%29%29
f%C2%A2%28x%29=%285x%5E2%2B56%29%2F%282sqrt%28x%2B14%29%29
equate to zero and solve for x
%285x%5E2%2B56%29%2F%282sqrt%28x%2B14%29%29=0-> will be zero if
5x%5E2%2B56=0
5x%5E2=-56
x%5E2=-56%2F5
x=sqrt%28-56%2F5%29-> complex solution, derivative is undefined
if %282sqrt%28x%2B14%29%29 derivative is undefined
Critical points are points where the function is defined and its derivative is zero or undefined

x=+-56%2F5, x=0, x=+-14-> a critical points
Identify critical points not in f (x ) domain : x%3E=+-14%7D%7D%0D%0A%0D%0A+domain+of+%7B%7B%7Bf%28x%29 is [+-14%7D%7D%2C%7B%7B%7Binfinity)
f%28x%29=x%5E2%2Asqrt%28x%2B14%29
f%28x%29=%28-56%2F5%29%5E2%2Asqrt%28-56%2F5%2B14%29
f%28x%29=209.9
maximum is at: (-14,0)
f%28x%29=x%5E2%2Asqrt%28x%2B14%29
f%28x%29=%28-14%29%5E2%2Asqrt%28-14%2B14%29
f%28x%29=%28-14%29%5E2%2Asqrt%280%29
f%28x%29=0
minimum is at: (-14,0)
f%28x%29=0%5E2%2Asqrt%280%2B14%29
f%28x%29=0
minimum is at: (-14,0) and (0,0)
maximum is at (-56%2F5,209.9)

f%28x%29 is monotone in intervals:
(-14,-56%2F5) -> f%28x%29 is increasing
(-56%2F5,0)->f%28x%29 is decreasing