Question 1024797
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Find the absolute extrema of 
f(x)=x^a(1-x)^b
where 0</= x =/< 1 and a and b are constants, both >1

I have the derivative 
f'(x)=ax^(a-1)(1-x)^b-x^(a)b(1-x)^(b-1)

Unsure of where to go from this point. Thank you for the help
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<pre>
Next step is to write an equation f'(x) = 0, which gives you this:

{{{ax^(a-1)(1-x)^b}}} = {{{x^(a)b(1-x)^(b-1)}}}.   (1)

Now cancel both sides of (1) by the factor {{{x^(a-1)*(1-x)^(b-1)}}}. You will get

a*(1-x) = b*x.

a - ax = bx

a = (a+b)*x,

x = {{{a/(a + b)}}}.   1-x = {{{b/(b+a)}}}.

f(x) = {{{(a/(a+b))^a}}}.{{{(b/(a+b))^b}}}
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