Question

Help please!!!
Find the derivative at each critical point and determine the local extrema values.
f(x)=x^2(sqrt(3-x))

Answer 1

this is an equation not a function
do you mean f(x) = x^2(sort(3 - x))

Answer 2

Yes. sorry for the typo. @cwrw238

Answer 3

fixed it!(:

Answer 4

use the product and the chain rules to find the derivative:
\[f'(x)=x^2*\frac{1}{2}(3-x)^{-\frac{1}{2}}*(-1)+2x\sqrt{3-x}\]

Answer 5

once I derive it what do I do next? @bahrom7893

Answer 6

to me, I do something like \[f(x) = x^2 \sqrt{3-x}=\sqrt{3x^4-x^5}=(3x^4-x^5)^{\frac{1}{2}}\]take derivative this f(x) I get
\[f'(x) = \dfrac{12x^3 -5x^4}{2\sqrt{3x^4-x^5}}\]then solve for x by letting numerator =0; combine with the condition of denominator not =0.

Answer 7

so 12x^3-5x^4 =0? then solve for x?

Answer 8

@Loser66

Answer 9

yes,

Answer 10

x=0 and x=12/5?

Answer 11

reject x =0 , because it make denominator =0 leads to the whole f'(x) undefined

Answer 12

okay.. that makes sense,

Answer 13

the condition for denominator not =0 is 0<x<3 and your critical point x = 12/5 satisfies that condition. BINGOOOOOO! go ahead, friend

Answer 14

Ohhhh... so how do i find if theres any local extrema values @Loser66

Answer 15

so far, you have 3 critical points x =0, x = 3 and x =12/5, right? plug them back to the original function to find out which f is biggest/ smallest . I mean f(0), f(3) and f (12/5)

Answer 16

the biggest valuse would be the local max right?

Answer 17

yup