Question

The maximum value of $f\left(x\right)=\left[x\right(-1)+1{]}^{\frac{1}{3}},0\le x\le 1$ is\\
a) ${\left(\frac{1}{3}\right)}^{\frac{1}{3}}$
b) $\frac{1}{2}$
c) 1
d) zero

Answer 1

Let $f\left(x\right)=\left[x\right(-1)+2{]}^{\frac{1}{3}},0\le x\le 1$
$=(x^2-x+1{)}^{\frac{1}{3}}$
On differentiating w.r.t x, we get
$f^{\prime }\left(x\right)=\frac{1}{3}(x^2-x+1{)}^{\frac{1}{3}-1}(2x-1)=\frac{1(2x-1)}{3(x^2-x+1{)}^{\frac{2}{3}}}$
Now, put $f^{\prime }\left(x\right)=0\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}ϵ[0,1]$
So, $x=\frac{1}{2}$ is a critical point.

Now, we evaluate the value of f at critical point $x=\frac{1}{2}$ and at the end points of the interval [0,1].
$\begin{array}{cc}Atx=0& f\left(0\right)=(0-0+1{)}^{\frac{1}{3}}=1\\ Atx=1& f\left(1\right)=(1-1+1{)}^{\frac{1}{3}}=1\\ Atx=\frac{1}{2},& f\left(\frac{1}{2}\right)={(\frac{1}{4}-\frac{1}{2}+1)}^{\frac{1}{3}}={\left(\frac{3}{4}\right)}^{\frac{1}{3}}\end{array}$
$\therefore$ Maximum value of f(x) is 1 at x=0, 1.
Hence, (c) is the correct option.