Answer
$f(1)=16$ absolute maximum, no absolute minimum.
Work Step by Step
Step 1. Given $f(x)=(7+x)(11-3x)^{1/3}$, we need to find its critical points and decide the extrema.
Step 2. Take the derivative to get $f'(x)=(11-3x)^{1/3}+\frac{1}{3}(7+x)(11-3x)^{-2/3}(-3)=(11-3x)^{-2/3}(11-3x-7-x)=(11-3x)^{-2/3}(4-4x)$
Step 3. Critical points happen when $f'(x)=0$, undefined, or at endpoints. We have critical points for this function at $x=1, \frac{11}{3}$, which gives $f(1)=(7+1)(11-3)^{1/3}=16$ and $f(\frac{11}{3})=(7+\frac{11}{3})(11-3(\frac{11}{3}))^{1/3}=0$
Step 4. This function has no end points; test end behaviors: as $x\to \pm\infty, f(x)\to -\infty$; thus there will be no absolute minimum. As $f(\frac{11}{3})\lt f(1)$, $f(\frac{11}{3})$ can not be an absolute maximum and will not be considered any further.
Step 5. Find the second derivative as $f''(x)=-4(11-3x)^{-2/3}-\frac{2}{3}(11-3x)^{-5/3}(-3)(4-4x)=(11-3x)^{-5/3}(-44+12x+8-8x))=4(11-3x)^{-5/3}(x-9)$.
Step 6. Evaluate $f''(1)=4(11-3)^{-5/3}(1-9)\lt0$, thus $f(1)$ is a local maximum.
Step 7. Because $f'\gt0$ when $x\lt1$ and $f'\lt0$ when $x\gt1$, we can further identify $f(1)=16$ as an absolute maximum over the domain
Step 8. See graph.
