함수의 증가와 감소

domain=$(-\infty,\infty)$

c.p. at $c=1$;

increasing on $(-\infty,\infty)$;

domain=$(-\infty,\infty)$

c.p. at $c=1$;

decreasing on $(1,\infty)$

increasing on $(-\infty,1)$;

rel. max at $x=1$.

domain=$(-\infty,-1)\cup(-1,1)\cup(1,\infty)$

c.p. at $c=0$;

decreasing on $(-\infty,-1)$ and $(-1,0)$;

increasing on $(0,1)$ and $(1,\infty)$;

rel. min at $x=0$;

domain=$(-\infty,-2)\cup(-2,4)\cup(4,\infty)$

no c.p.;

decreasing on entire domain, $(-\infty,-2)$, $(-2,4)$ and $(4,\infty)$

domain=$(-\infty,0)\cup(0,\infty)$;

c.p. at $c=2,6$;

decreasing on $(-\infty,0)$, $(0,2)$ and $(6,\infty)$;

increasing on $(2,6)$;

rel. min at $x=2$; rel. max at $x=6$.

domain=$(-\infty,\infty)$

c.p. at $c=-3\pi/4,-\pi/4,\pi/4,3\pi/4$;

decreasing on $(-3\pi/4,-\pi/4)$ and $(\pi/4,3\pi/4)$;

increasing on $(-\pi,-3\pi/4)$, $(-\pi/4,\pi/4)$ and $(3\pi/4,\pi)$;

rel. min at $x=-\pi/4,3\pi/4$;

rel. max at $x=-3\pi/4,\pi/4$.

domain = $(-\infty,\infty)$;

c.p. at $c=-1,1$;

decreasing on $(-1,1)$;

increasing on $(-\infty,-1)$ and $(1,\infty)$;

rel. min at $x=1$;

rel. max at $x=-1$

Functions $A(x)$ and $B(x)$ share something in common that sets them apart from the others: they have a horizontal tangent line only once. In particular, $A'(-2) \neq 0$ and $B'(2) \neq 0$. The only listed functions that do not have two distinct roots are $l(x)$ and $p(x)$. Since $l(-2) \neq 0$ and $p(2) \neq 0$, we conclude $$A'(x)=l(x) \qquad B'(x)=p(x)$$

Function $C(x)$ is never decreasing. Its tangent line is horizontal when $x = \pm 2$, but the curve never decreases, so $C'(x) \geq 0$ for all $x$ and $C'(2)=C'(-2)=0$. The only function that matches this is $n(x)=(x-2)^2(x+2)^2$. Since its linear terms have even powers, it is never negative, and its roots are precisely $x=\pm 2$. $$C'(x)=n(x)$$

For the functions $D(x)$ and $E(x)$ we consider their behaviour near $x=0$. $D(x)$ is decreasing near $x=0$, so $D'(0)<0$, which matches with $o(0)<0$. Contrastingly, $E(x)$ is increasing near zero, so $E'(0)>0$, which matches with $m(0)>0$. $$D'(x)=o(x)\qquad E'(x)=m(x)$$

The domain of $f(x)$ is all real numbers except $-3$ (because when $x=-3$ the denominator is zero). For $x\neq -3$, we differentiate using the quotient rule: $$f'(x)=\frac{e^x(x+3) - e^x(1)}{(x+3)^2} = \frac{e^x}{(x+3)^2} (x+2)$$ Since $e^x$ and $(x+3)^2$ are positive for every $x$ in the domain of $f(x)$, the sign of $f'(x)$ is the same as the sign of $x+2$. We conclude that $f(x)$ is increasing for every $x$ in its domain with $x+2>0$. That is, over the open interval $(-2,\infty)$.

Since we can't take the square root of a negative number, $f(x)$ is only defined when $x \ge 1$. Furthermore, since we can't have zero as a denominator, $x=-2$ is not in the domain — but as long as $x \ge 1$, we also have $x \ne -2$. So, the domain of the function is $[1,\infty)$.

In order to find where $f(x)$ is increasing, we find where $f'(x)$ is positive. $$f'(x)=\frac{\frac{2x+4}{2\sqrt{x-1}}-2\sqrt{x-1}}{(2x+4)^2} =\frac{(x+2)-2(x-1)}{\sqrt{x-1}(2x+4)^2} =\frac{-x+4}{\sqrt{x-1}(2x+4)^2}$$ The denominator is never negative, so $f(x)$ is increasing when the numerator of $f'(x)$ is positive, i.e. when $4-x>0$, or $x<4$. Recalling that the domain of definition for $f(x)$ is $[1,+\infty)$, we conclude that $f(x)$ is increasing on the open interval $(1,4)$.

The domain of arctangent is all real numbers. The domain of the logarithm function is all positive numbers, and $1+x^2$ is positive for all $x$. So, the domain of $f(x)$ is all real numbers.

In order to find where $f(x)$ is increasing, we find where $f'(x)$ is positive. $$f'(x)= \frac 2 {1+x^2} - \frac{2x}{1+x^2} = \frac{2-2x}{1+x^2}$$ Since the denominator is always positive, $f(x)$ is increasing when when $2-2x>0$. We conclude that $f(x)$ is increasing on the open interval $(-\infty,1)$.