7.2 First derivative - increasing or decreasing

For each of the graphs, consider where the derivative is positive, negative, and zero.

$\textcolor{green}{A'(x)=l(x)} \qquad \textcolor{blue}{B'(x)=p(x)} \qquad \textcolor{red}{C'(x)=n(x)} \qquad \textcolor{orange}{D'(x)=o(x)}\qquad \textcolor{purple}{E'(x)=m(x)}$

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

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 $x=\pm 2$.

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 $m(0)>0$.

Where is $f'(x)>0$?

$(-2,\infty)$

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:

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 $(-2,\infty)$.

Consider the signs of the numerator and the denominator of $f'(x)$.

$(1,4)$

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.

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 $(1,4)$.

Remember $\ds\diff{}{x}\{\arctan x\}=\dfrac{1}{1+x^2}$.

$(-\infty,1)$

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.

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 $(-\infty,1)$.