Minimum of the real valued function \[{\rm{f}}\left( {\rm{x}} \right) = {\left( {{\rm{x}} – 1} \right)^{\frac{2}{3}}}\] occurs at x equal to A. \[ – \infty \] B. 0 C. 1 D. \[\infty \]

” option2=”0″ option3=”1″ option4=”\[\infty \]” correct=”option1″]

The minimum of the real valued function $f(x) = \left(x – 1\right)^{\frac{2}{3}}$ occurs at $x=1$.

To see this, we can first find the derivative of $f$. The derivative of $f$ is $f'(x) = \frac{2}{3}(x-1)^{-\frac{1}{3}}$.

We can then set $f'(x)=0$ and solve for $x$. This gives us $x=1$.

We can also see that $f(x)$ is increasing for $x<1$ and decreasing for $x>1$. This means that the minimum value of $f(x)$ must occur at $x=1$.

Finally, we can graph $f(x)$ to see that the minimum value does indeed occur at $x=1$.

[asy]
unitsize(1 cm);

draw((-2,0)–(2,0));
draw((0,-1)–(0,2));

real ticklen=1;
real tickspace=1;
real axisarrowsize=0.14inch;

real ticklength=0.14inch;
real vectorarrowsize=0.2inch;

pen axispen=black+1.2bp;
pen tickpen=black+0.8bp;
pen tickdown=black+0.8bp;
pen tickdown=black+0.8bp;
pen vectorpen=black+1.2bp;

real i;

for (i=-2; i<=2; i+=0.2) {
draw((i,-0.1)–(i,0.1));
}

for (i=-1; i<=1; i+=0.2) {
draw((0,i)–(0.1,i));
}

label(“$x$”, (2,0), E);
label(“$y$”, (0,2), N);

draw((0,0)–(1,1));

draw((0,-0.1)–(0,0.1),Arrows);
draw((1,-0.1)–(1,0.1),Arrows);

label(“$1$”, (1,0), S);

real f(real x) {
return (x-1)^(2/3);
}

draw(graph(f,-2,2),red);
[/asy]