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Section 3.1 #12: Find the critical numbers of the function $g(x)=x^4-8x^2$.
Solution: First differentiate and set the derivative equal to zero:
$$g'(x)=4x^3-16x \stackrel{\rm{set}}{=} 0.$$ To solve the equation $4x^3-16x=0$ factor an $x$ out of the left-hand side to get $$x(4x^2-16)=0.$$ Using the zero-product property of algebra, we see that $x=0$ or $4x^2-16=0$. To solve the second equation we see $4x^2=16$, or $x^2=4,$ or $x=\pm 2$. We have found three critical numbers: $x=-2,0,2$.

Section 3.1 #25: Find the absolute extrema of the function $g(x)=\dfrac{t^2}{t^2+3}$ on the interval $[-1,1]$.
Solution: Compute the derivative and set it equal to zero: $$g'(x) = \dfrac{(t^2+3)(2t)-t^2 (2t)}{(t^2+3)^2}=\dfrac{2t^3+6t-2t^3}{(t^2+3)^2} = \dfrac{6t}{(t^2+3)^2} \stackrel{\mathrm{set}}{=} 0.$$ Take note that there are no real numbers $t$ for which $t^2+1=0$ (if we found one, it would also be on our list of critical points). So we may now solve the equation $\dfrac{6t}{(t^2+3)^2}=0$ by multiplying by $t^2+3$ to get $6t=0$ which has solution $t=0$, which is our only critical point. Now to find the absolute extrema, we must evaluate $g$ at the critical point $t=0$ and the endpoints $t=-1$ and $t=1$:
So compute $$g(-1)=\dfrac{(-1)^2}{(-1)^2+3} = \dfrac{1}{1+3} = \dfrac{1}{4},$$ $$g(0)=\dfrac{0^2}{0^2+3}=0,$$ and $$g(1)=\dfrac{1^2}{1^2+3}=\dfrac{1}{4}.$$ The largest value, $\dfrac{1}{4}$, says we have an absolute maximum of $\dfrac{1}{4}$ at the $x$-values $-1$ and $1$. The smallest value, $0$, says we have an absolute minimum of $0$ at the $x$-value $0$.

Section 3.1 #33: Find the absolute extrema of the function $f(x)=\sin(x)$ on the interval $\left[ \dfrac{5\pi}{6}, \dfrac{11 \pi}{6} \right]$.
Solution: First differentiate the function and set its derivative equal to zero:
$$\cos(x) \stackrel{\mathrm{set}}{=}0.$$ (Also note that $\cos(x)$ is not undefined for any $x$). Basic trigonometry tells us that the solution of this equation is $x= \dfrac{\pi}{2} + 2\pi k$ or $x=\dfrac{3\pi}{2} + 2\pi k$ for any integer $k$. Let us make a table of some of the solutions for various $k$:
$k=$$\frac{\pi}{2}+2\pi k=$$\frac{3\pi}{2}+2\pi k=$
$-1$$-\frac{3\pi}{2}=-\frac{9\pi}{6}$$-\frac{\pi}{2}=-\frac{3\pi}{6}$
$0$$\frac{\pi}{2}=\frac{3\pi}{6}$$\frac{3\pi}{2}=\frac{9\pi}{6}$
$1$$\frac{5\pi}{2}=\frac{15\pi}{6}$$\frac{7\pi}{2}=\frac{21\pi}{6}$
We see the only critical point in the interval is $\dfrac{9\pi}{6}=\dfrac{3\pi}{2}$. So we must test this critical point and the endpoints of the interval into the function $f$: compute $$f \left(\dfrac{5\pi}{6} \right)=\sin \left( \dfrac{5\pi}{6} \right)=\dfrac{1}{2},$$ $$f \left( \dfrac{11\pi}{6} \right)=\sin \left( \frac{11\pi}{6} \right) =-\dfrac{1}{2},$$ and $$f \left( \dfrac{9\pi}{6} \right)=\sin \left( \dfrac{9\pi}{6} \right)=\sin\left(\dfrac{3\pi}{2} \right)=-1.$$ The absolute maximum is $\dfrac{1}{2}$ and it occurs at $x=\dfrac{5\pi}{6}$. The absolute minimum is $-1$ and it occurs at $x=\dfrac{3\pi}{2}$.

Section 3.2 #15: Determine whether Rolle's theorem can be applied on the interval $[-1,3]$ for the function $$f(x)=\dfrac{x^2-2x-3}{x+2}.$$ If it can be applied, find the value(s) of $c$ such that $c$ lies in the interval $[-1,3]$ and $f'(c)=0$.
Solution: The function $f$ is continuous and differentiable at all real numbers except at $x=-2$ (it causes division by zero). However, $-2$ is not in the interval $[-1,3]$, and since $f(-1)=\dfrac{1+2-3}{-1}=0$ and $f(3)=\dfrac{9-6-3}{5}=0$, we may apply Rolle's theorem. This means there exists at least one $c$ between $-1$ and $3$ such that $f'(c)=0$. So compute the derivative of $f$: $$\begin{array}{ll} f'(x)&=\dfrac{(x+2)(2x-2)-(x^2-2x-3)(1)}{(x+2)^2} \\ &=\dfrac{2x^2-2x+4x-4-x^2+2x+3}{(x+2)^2} \\ &=\dfrac{x^2+4x-1}{(x+2)^2}. \end{array} $$ To find the requested values of $c$, set the derivative equal to zero: $$\dfrac{x^2+4x-1}{(x+2)^2} = 0.$$ Since we are not considering $x= -2$, we may multiply by $(x+2)^2$ and get $$x^2+4x-1=0.$$ Solve this equation using the quadratic formula: $$x=\dfrac{-4 \pm \sqrt{16-4(1)(-1)}}{2} = -2 \pm \dfrac{1}{2}\sqrt{20} = -2\pm \sqrt{5}.$$

Since we are restricted to the interval $[-1,3]$ we must throw out the solution $-2 - \sqrt{5}$, leaving us the value $c=-2+\sqrt{5}$ (which is of course in the interval, since $\sqrt{5}$ is a little bit larger than $2$).

Section 3.2 #46: Determine whether the mean value theorem can be applied to $f$ on the closed interval $[a,b]$. If the mean value theorem can be applied, find all values of $c$ in the open interval $(a,b)$ such that $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.
$$f(x)=\cos(x)+\tan(x), \quad [0,\pi].$$ Solution: The function $f$ is *not* continuous at $x=\dfrac{\pi}{2}$ (which lies in the interval $[0,\pi]$) because $\tan \left( \dfrac{\pi}{2} \right)$ is undefined. Therefore we may not apply the mean value theorem on this problem to this function to $f$ on the interval $[0,\pi]$.