The maximum value of\( [x(x-1)+1]^{\frac{1}{3}},0≤x≤1\) is
\((\frac{1}{3})^{\frac{1}{3}}\)
\(\frac{1}{2}\)
1
0
Let \(ƒ(x)=[x(x-1)+1]\frac{1}{3}.\)
\(∴ƒ'(x)=\frac{2x-1}{3[x(x-1)+1]\frac{2}{3}}\)
Now\(,ƒ'(x)=0⇒x=\frac{1}{2}\)
Then,we evaluate the value of f at critical point \(\frac{1}{2}\) and at the end points of the
interval \([0,1]\){i.e.,at \(x=0 and x=1\)}.
\(ƒ(0)=[0(0-1)+1]\frac{1}{3}=1\)
\(ƒ(1)=[1(1-1)+1]\frac{1}{3}=1\)
\(ƒ(\frac{1}{2})=[\frac{1}{2}(-\frac{1}{2})+1]^{\frac{1}{3}}=(\frac{3}{4})^{\frac{1}{3}}\)
Hence,we can conclude that the maximum value of \(f\) in the interval \([0,1]is 1.\)
The correct answer is C.
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