Question 1155624
Determine the location and value of the absolute extreme values of f on the given
interval, if they exist. f(x)=12x^(2/3) -x on [0,1728]
<pre>
{{{f(x)=12x^(2/3) -x}}} on [0,1728] 

{{{matrix(2,1,"","f'(x)"=(2/3)(12)x^(2/3-1)-1=8x^(-1/3)-1)}}}

To find potential extrema set f'(x) = 0 

{{{matrix(5,1,
8^""/x^("1/3") - 1=0,
8/x^("1/3") = 1,
8=x^(1/3),
8^3=x,
512=x)}}}

2nd derivative test for max or min

{{{matrix(2,1,"","f'(x)"=8x^(-1/3)-1)}}}

{{{matrix(2,1,"","f''(x)"=expr(-8/3)x^(-1/3-1)=expr(-8/3)x^(-4/3))}}}

{{{matrix(2,1,"","f''(512)"=expr(-8/3)512^(-1/3-1)=expr(-8/3)512^(-4/3)=-6.5*10^(-4))}}}

That's negative so relative maximum at x=512.

We find the value there.

{{{f(512)=12*512^(2/3) -512=256}}}

So relative maximum is (512,256)

We must examine to see if endpoints of interval are higher points.

{{{f(0)=12*0^(2/3) -0=0}}}
{{{f(1728)=12*1728^(2/3) -1728=0}}}

Thus absolute maximum at (512,256), absolute minima at (0,0) and (1728,0)

Edwin</pre>