Answer
$$
f(x)=\ln\frac{5x^{2}+4}{x^{2}+1}
$$
(a)
The critical numbers : $x=0$.
b)
The function is increasing , on interval $(0, \infty)$
(c)
The function is decreasing on interval $(-\infty, 0)$ .
Work Step by Step
$$
f(x)=\ln\frac{5x^{2}+4}{x^{2}+1}
$$
We find $y^{\prime}(x) $ first, using the power rule and the chain rule.
$$
\begin{aligned}
f^{\prime}(x) &=\frac{d}{dx}[\ln\frac{5x^{2}+4}{x^{2}+1}]\\
&=\frac{d}{dx}[\ln(5x^{2}+4)- \ln(x^{2}+1)]\\
&=\frac{10 x}{5 x^{2}+4}-\frac{2 x}{x^{2}+1} \\
&=\frac{10 x\left(x^{2}+1\right)}{\left(5 x^{2}+4\right)\left(x^{2}+1\right)}-\frac{2 x\left(5 x^{2}+4\right)}{\left(5 x^{2}+4\right)\left(x^{2}+1\right)} \\
&=\frac{10 x^{3}+10 x-10 x^{3}-8 x}{\left(5 x^{2}+4\right)\left(x^{2}+1\right)} \\
&=\frac{2 x}{\left(5 x^{2}+4\right)\left(x^{2}+1\right)}
\end{aligned}
$$
To find the critical numbers, we first find any values of $x$ that make $f^{\prime}(x)=0 $ , here $f^{\prime}(x)=0 $ when $x=0. $ Next, we find any values of $x$ where $f^{\prime}(x) $ fails to exist. This occurs whenever the denominator of $f^{\prime}(x) $ is $0$, here we observe that the denominator $\left(5 x^{2}+4\right)\left(x^{2}+1\right)$ does not equal to 0. So, $x=0 $ is the only critical number.
(a)
The critical numbers : $x=0$.
We can still apply the first derivative test, however, to find where $f$ is increasing and decreasing. We observe, the number $0$ divides the number line into two intervals: $(-\infty, 0)$ and $(0, \infty)$. use a test point in each of these intervals to find that sign of $f^{\prime} $ (This can also be determined by observing that $f^{\prime}(x) $ is the quotient of $2x$, which is negative or positive , and $\frac{2 x}{\left(5x^{2}+4\right)\left(x^{2}+1\right)}$ , which is always positive.) This means that the function $f $ is decreasing on $(-\infty, 0)$ and is increasing on$(0, \infty)$.
So,
b)
The function is increasing , on interval $(0, \infty)$
(c)
The function is decreasing on interval $(-\infty, 0)$ .
