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Which of the following x-coordinates is a candidate for being an extreme value for the function f(x)=x^2 2^x a) -1, b) 2, c) 1, d) 0
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\[f(x)=x^2 2^x\]
The derivative is f'(x)=2x2^xln2, I don't know where to go from there
find the critical values
by equating the derivative to 0
0=2x2^x ln2 0=2x2^x is that right so far?
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yep
Ok now i'm not too sure what to do. would I remove the 2 from 2x?
Oh wait would I do 2x=0 then x=0. Then I sub in the x=0 into 2^x?
wait
are u sure abt ur derivative
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I'm not too sure but I think it's right
k
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