[amp_mcq option1=”\\[ – \\infty \\]” option2=”0″ option3=”1″ option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n
The minimum of the real valued function $f(x) = \\left(x – 1\\right)^{\\frac{2}{3}}$ occurs at $x=1$.<\/p>\n
To see this, we can first find the derivative of $f$. The derivative of $f$ is $f'(x) = \\frac{2}{3}(x-1)^{-\\frac{1}{3}}$.<\/p>\n
We can then set $f'(x)=0$ and solve for $x$. This gives us $x=1$.<\/p>\n
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 $x=1$.<\/p>\n
Finally, we can graph $f(x)$ to see that the minimum value does indeed occur at $x=1$.<\/p>\n
[asy]
\nunitsize(1 cm);<\/p>\n
draw((-2,0)–(2,0));
\ndraw((0,-1)–(0,2));<\/p>\n
real ticklen=1;
\nreal tickspace=1;
\nreal axisarrowsize=0.14inch;<\/p>\n
real ticklength=0.14inch;
\nreal vectorarrowsize=0.2inch;<\/p>\n
pen axispen=black+1.2bp;
\npen tickpen=black+0.8bp;
\npen tickdown=black+0.8bp;
\npen tickdown=black+0.8bp;
\npen vectorpen=black+1.2bp;<\/p>\n
real i;<\/p>\n
for (i=-2; i<=2; i+=0.2) {
\n draw((i,-0.1)–(i,0.1));
\n}<\/p>\n
for (i=-1; i<=1; i+=0.2) {
\n draw((0,i)–(0.1,i));
\n}<\/p>\n
label(“$x$”, (2,0), E);
\nlabel(“$y$”, (0,2), N);<\/p>\n
draw((0,0)–(1,1));<\/p>\n
draw((0,-0.1)–(0,0.1),Arrows);
\ndraw((1,-0.1)–(1,0.1),Arrows);<\/p>\n
label(“$1$”, (1,0), S);<\/p>\n
real f(real x) {
\n return (x-1)^(2\/3);
\n}<\/p>\n
draw(graph(f,-2,2),red);
\n[\/asy]<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[ – \\infty \\]” option2=”0″ option3=”1″ option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20239","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n