Find the absolute max and minimum values of x^2*(x-7)^4/5

[boshiboshibom]

Here is the question:
Find the absolute max and minimum values of (x^2)*[(x-7)^4/5], 0<x<orequal to 7

So far I have done the derivative, which is the product rule and chain(?) rule.

2x(x-7)^4/5+x^2*{4/5[(x-7)^-1/5]*1}

and I think solve for f(x)=0? so


2x(x-7)^4/5+x^2*{4/5[(x-7)^-1/5]*1}=0

now I am not sure how to proceed because I can't find a common factor to pull out. Please help and explain how to arrive at the answer, thank you.

Edit: Added the grouping symbols, sorry, forgot to to add them in copy and pasting of function

 

[Deleted member 4993]

Re: Find the absolute max and minimum values of x^2*(x-7)^4/

Here is the question:
Find the absolute max and minimum values of x^2*(x-7)^4/5, 0<x<orequal to 7

So far I have done the derivative, which is the product rule and chain(?) rule.

2x(x-7)^4/5+x^2*{4/5[(x-7)^-1/5]*1}

and I think solve for f(x)=0? so


2x(x-7)^4/5+x^2*{4/5[(x-7)^-1/5]*1}=0

now I am not sure how to proceed because I can't find a common factor to pull out. Please help and explain how to arrive at the answer, thank you.

\(\displaystyle 2x(x-7)^{\frac{4}{5}} \ + \ x^2\frac{4}{5}\cdot \frac{1}{(x-7)^{\frac{1}{5}}} \ = \ 0\)

Multiplying by (x-7)[sup:29cngne9]1/5[/sup:29cngne9] we get:

\(\displaystyle 2x(x-7) \ + \ x^2\frac{4}{5} \ = \ 0\)

Now continue.....

 

[lookagain]

Re: Find the absolute max and minimum values of x^2*(x-7)^4/

Here is the question:
Find the absolute max and minimum values of \(\displaystyle >>>\)x^2*(x-7)^(4/5) \(\displaystyle <<< \\), 0 < x <or equal to 7

.

2x(x-7)^(4/5) + x^2*(4/5)[(x-7)^(-1/5)]*1}

2x(x-7)^(4/5) + x^2*{4/5[(x-7)^(-1/5)]*1}=0

OP, put grouping symbols around your fractions as shown in this quote box, and you
must have them for fractions that are exponents because of the order of operations.