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OpenStudy (anonymous):
Help Please! Find the global extreme values of the function f on the given interval and where they occur. f(x)=x^(2/3)*(x-5)^2 on the interval [0,2].
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OpenStudy (amistre64):
test the end points, then find your critical values of the derivative to test
OpenStudy (anonymous):
I got \[f \prime (x)= 2(x-5)x ^{2/3} + \frac{ 2(x+5)^2 }{ 3x^{1/3} }\] I equaled that to zero and got \[x=5 \] and \[x=\frac{ 5 }{ 4 }\] then I pluged the values in and got f(0)=0 , f(5)=0 and f(5/4)=69.42..... @amistre64 I'm guessing my mins are at f(0) and f(5)? and my max at f(5/4) But how would I prove my point?
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