Question

all critical points of 3x(8-x)^3

Answer 1

Critical points means the derivative is 0,
You can start by differentiating that then set it equals to zero

Answer 2

I don't know what to do with the cubed part

Answer 3

Product and chain rule here,
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Product rule
\[\frac{d}{dx}(uv)=udv+vdu\]
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\[y'=(3x)(3)(8-x)^2(-1)+(8-x)^3(3)\]
From chain rule you get (-1) for that derivation.
\[y'=-12 (x-8)^2 (x-2)\]

Answer 4

Thank you!

Answer 5

After that set y'=0 will give you the critical points
-12 (x-8)^2 (x-2)=0

Answer 6

so x= 8 and x=2

Answer 7

Yes