# Find the interval on which the function f(x)=(x-3)^2(x+3)^2 is increasing and decressing?

Sketch the graph of y=f(x) and identify any local nmaxima and minima. Any Global extrema should also be identified.

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- 1 decade agoFavorite Answer
f(x)=[(x-3)(x+3)]^2=(x^2-3^2)^2=(x^2-9)^2=x^4-18x^2+81

x^2=y

f(y)=y^2-18y^2+81-ec of second grade

a=1,b=-18,c=81

delta=b^2-4ac=324-324=0

f-constant

- 1 decade ago
The sketching part i'm unable to do it online...

To identify local maxima and minima, do it by differentiation...

y= (x-3)^2(x+3)^2

dy/dx=4x(x-3)(x+3)

Let dy/dx=0, this is where turning points occur.

0=4x(x-3)(x+3)

x=0 or x=3 or x=-3

Next, the check whether it is max or min pt, use the 2nd derivative test.

d2y/dx2=12(x^2-3)

Substitute each of the x values into the eqn

when x=0, d2y/dx2=-ve(Max point)

when x=3, d2y/dx2= +ve (Min pt)

when x=-3, d2y/dx2 = +ve(Min pt)

Hope this helps..:)

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