On critical points

The domain of a logarithmic function requires its argument to be positive:

$$\frac{x}{x^2+1}>0$$

Since the denominator $x^2+1$ is always positive, the domain of $f$ is $x>0$. Only critical points within this domain are valid.

Before differentiating, $f(x)$ can be rewritten using log properties:

$$f(x)=\ln (x)-\ln (x^2+1)$$

Then, differentiating:

$$f^{\prime }(x)=\frac{1}{x}-\frac{2x}{x^2+1}$$

$f^{\prime }(x)$ is undefined when $x=0$. However, since $f(0)$ is also undefined, $x=0$ is not a critical point.

Next, solve $f^{\prime }(x)=0$.

$$\begin{array}{rl}\frac{1}{x}-\frac{2x}{x^2+1}& =0\\ \frac{2x}{x^2+1}& =\frac{1}{x}\\ 2x^2& =x^2+1\\ x^2& =1\\ x& =1,-1\end{array}$$

Checking these in the original function,

• $f(-1)$ is undefined.

• $f(1)$ is defined.

So the only critical point is $x=1$.

Because the domain of $f(x)$ is $x>0$, we only need to test intervals to the right of $0$ for the sign of $f^{\prime }(x)$.

At $x=1$, $f^{\prime }(x)$ changes from positive to negative, meaning $f$ has a relative maximum at $x=1$.

The derivative is

$$\begin{array}{rl}{f}^{\prime }(x)& =x{e}^{1/x}(-\frac{1}{x^2})+e^{1/x}(1)\\ & =e^{1/x}(1-\frac{1}{x})\end{array}$$

$f^{\prime }(x)$ is undefined when $x=0$. But $f(x)$ is also undefined at $x=0$, so $x=0$ is not a critical point.

Next, find where $f^{\prime }(x)=0$, which occurs when either

• $e^{1/x}=0$, or

• $1-\frac{1}{x}=0$

$e^{1/x}$ is always positive and never equals $0$.

So $f^{\prime }(x)=0$ when

$$\begin{array}{rl}1-\frac{1}{x}& =0\\ \frac{1}{x}& =1\\ x& =1\end{array}$$

$x=1$ is the only critical point, and $x=0$ is a point of discontinuity.

Both $x$-values are included on the sign chart.

At $x=0$, $f(x)$ is not continuous. Even though $f^{\prime }(x)$ changes from positive to negative across $0$, that change does not create a relative maximum or minimum because the function isn’t defined there.

At $x=1$, $f^{\prime }(x)$ changes from negative to positive, so $f$ has a relative minimum at $x=1$.

Differentiating,

$$f^{\prime }(x)=-\frac{1}{(x-1{)}^2}$$

Because the numerator is a constant, there is no value of $x$ where $f^{\prime }(x)=0$.

When $x=1$, $f^{\prime }(x)$ is undefined. But since $f(x)$ is also undefined at $x=1$, it is not a critical point.

So this function has no critical points. That’s fine - not every function has critical points.

Even without critical points, we can still use a sign chart to describe how the function behaves around the point of discontinuity. This idea will be important in section 5.4 on curve sketching.

We can reason directly from $f^{\prime }(x)$ instead of plugging in test points. The denominator $(x-1{)}^2$ is always positive (except where it’s undefined), and the negative sign in front makes the whole derivative negative. So

$$f^{\prime }(x)<0\text{ for all }x\ne 1$$

The function is decreasing for all $x$. There are no critical points, so there are no relative extrema. The vertical asymptote at $x=1$ means the function approaches $\infty$ and $-\infty$ on either side, so there is no absolute maximum or minimum either.