Extreme Values and the Mean Value Theorem

The domain of $g$ is every real number, since $x^{1/3}$ is defined for all $x$.

$g'(x)=\frac43x^{1/3}-\frac43x^{-2/3}=\frac43x^{-2/3}(x-1)=\frac{4(x-1)}{3x^{2/3}}$

The numerator vanishes at $x=1$. The denominator vanishes at $x=0$, so $g'(0)$ does not exist — and $0$ is in the domain of $g$, which makes it a critical number too.

Check at $x=8$: $\frac{4(7)}{3(4)}=\frac73$, and directly $\frac43(2)-\frac43(\frac14)=\frac73$. ✓

$f$ is a polynomial, so it is continuous on this closed bounded interval and the EVT applies.

$f'(x)=3x^2-6x=3x(x-2)=0\ \Longrightarrow\ x=0,\ 2$

Both lie in $\left(-\frac12,4\right)$, and $f'$ never fails to exist. Evaluate at the critical numbers and both endpoints:

$f\!\left(-\tfrac12\right)=\tfrac18,\qquad f(0)=1,\qquad f(2)=-3,\qquad f(4)=17$

The largest of the four is $17$, the smallest is $-3$.

$4-x^2\ge0$ exactly on $[-2,2]$, where $f$ is continuous, so the EVT applies.

$f'(x)=\sqrt{4-x^2}+x\cdot\frac{-x}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}$

$f'(x)=0$ gives $x^2=2$, so $x=\pm\sqrt2$, both interior. ($f'$ also fails at $x=\pm2$, but those are the endpoints, which are tested anyway.)

$f(-2)=0,\quad f(-\sqrt2)=-2,\quad f(\sqrt2)=2,\quad f(2)=0$

Check: $f$ is odd, so the two extreme values must be negatives of each other. ✓

$f$ is rational and undefined only at $x=-2$, which is outside $[1,4]$; so $f$ is continuous on $[1,4]$ and differentiable on $(1,4)$.

Secant slope: $f(1)=\frac13$ and $f(4)=\frac23$, so $\dfrac{f(4)-f(1)}{4-1}=\dfrac{1/3}{3}=\dfrac19$.

$f'(x)=\frac{(x+2)-x}{(x+2)^2}=\frac{2}{(x+2)^2}=\frac19\ \Longrightarrow\ (c+2)^2=18$

So $c=-2\pm3\sqrt2$. Reject $c=-2-3\sqrt2\approx-6.24$; only $c=3\sqrt2-2\approx2.243$ lies in $(1,4)$.

(a) The EVT needs a closed bounded interval, and $(0,3]$ omits its left endpoint, so the guarantee is void. Indeed $\frac1x\to\infty$ as $x\to0^+$: no maximum. The minimum survives by luck — $f$ is decreasing, so its least value is $f(3)=\frac13$. The EVT is sufficient, not necessary.

(b) $h'(x)=-\dfrac{2}{3x^{1/3}}$ is never zero, and it does not exist at $x=0$, a point of $(-1,1)$. Rolle's differentiability hypothesis fails, so the theorem claims nothing: the graph has a cusp peak at $(0,1)$, not a horizontal tangent.

The hypotheses are given, so the MVT supplies a $c$ in $(2,6)$ with

$f(6)-f(2)=f'(c)\,(6-2)=4f'(c)$

Since $2\le f'(c)\le4$ and the multiplier $4$ is positive, $8\le f(6)-f(2)\le16$. Adding $f(2)=5$:

$13\le f(6)\le21$

Both ends occur: $f(x)=5+2(x-2)$ satisfies every hypothesis and gives $f(6)=13$, while $f(x)=5+4(x-2)$ gives $f(6)=21$.