<PRE><B>Question 1119676</B>
&lt;br&gt;
k(x) = {{{(2/3)x + (x+1)^(2/3)}}}&lt;br&gt;
k&#39;(x) = {{{2/3 + (2/3)(x+1)^(-1/3) = 2/3 + 2/(3(x+1)^(1/3))}}}&lt;br&gt;
(1) Set k&#39;(x) = 0 to look for local extrema.&lt;br&gt;
{{{2/(3(x+1)^(1/3)) = -2/3}}}
{{{1/(x+1)^(1/3) = -1}}}
{{{1/(x+1) = -1}}}   [by cubing both sides of the equation]
{{{1 = -x-1}}}
{{{x = -2}}}&lt;br&gt;
{{{k(-2) = -4/3 + (-1)^(2/3) = -4/3 + 1 = -1/3}}}&lt;br&gt;
The point (-2, -1/3) is a local extremum.  To see if it is a local maximum or a local minimum, check the second derivative.&lt;br&gt;
k&#39;&#39;(x) = {{{(-2/9)(x+1)^(-4/3)}}}&lt;br&gt;
k&#39;&#39;(-2) = {{{(-2/9)(-1)^(-4/3) = -2/9}}}&lt;br&gt;
The second derivative is negative at x = -2, so (-2,-1/3) is a local maximum.&lt;br&gt;
(2) Other local extrema can occur where the first derivative is undefined.&lt;br&gt;
For this function, the derivative is undefined at x = -1.  For x slightly less than -1 (more negative than -1), the derivative is large negative; for x slightly more than -1 (less negative than -1), the derivative is large positive.&lt;br&gt;
Since the derivative changes suddenly at x = -1 from large negative to large positive, there is a local minimum at x = -1, as long as the function is defined there.&lt;br&gt;
The function is defined at x = -1:&lt;br&gt;
k(-1) = {{{(2/3)(-1)+(-1+1)^(2/3) = -2/3}}}&lt;br&gt;
So the point (-1,-2/3) is a local minimum.&lt;br&gt;
(3) For absolute extrema, we need to check the function values at the endpoints of the given domain.&lt;br&gt;
Inspection shows that the derivative is always positive for x less than -2; that means as x goes to negative infinity there is no absolute minimum value.&lt;br&gt;
For the absolute maximum value, inspection again shows that the derivative is always positive for x greater than -1.  That means the absolute maximum is at the right endpoint of the domain.&lt;br&gt;
So the absolute maximum is (0,k(0)) = (0,1).</PRE>