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Posted by admin on 21st January and posted in Calculus
Question by Des: ???CAlculus???
Find all Critical numbers of the function f(x)=(9-x^2)^3/5
Best answer:
Answer by shadowca1964
1 set (9-x^2)^3/5 equal to zero then raise both sides to 5/3 getting (9-x^2)=0 so x=0 at x= +/- 3
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i am taking calculus too and wow, it feels like i jumped in the deep end w/o knowing fully how to swim.
that means to find all the points where the first derivative of that function is 0.
f ‘(x) = (3/5)[(9-x^2)^(-2/5)](-2x)
simplifies:
f ‘(x) = [-6x]/[5(9-x^2)^(2/5)
if we set that equal to 0, then the top of the fraction is all that we need to be 0
so
-6x=0
x = 0 is our critical point
To find critical numbers, you want to set the derivative equal to zero.
f ‘ (x) = (3/5)(9-x^2)^(-2/5)(-2x) = 0
Thus, -2x = 0 OR (9-x^2)^(-2/5) = 0.
But (9-x^2)^(-2/5) can NEVER equal 0 since it has a negative exponent.
So, the only critical number is x=0.
Critical numbers are where f’(x) = 0.
f’(x) = (3/5)(9-x^2)^(-2/5)*-2x = (-6x/5)(9-x^2)^(/2-5).
This equals 0 when either -6x/5 = 0 or when (9-x^2)^(-2/5) = 0.
If -6x/5 = 0, x = 0. If
(9-x^2)^(-2/5) = 0. Raise both sides to the (-5/2).
9 – x^2 = 0.
x^2 = 9.
x = 3 or x = -3.
Your critical points are x = -3, x = 0, and x = 3.