Sketching a function with logarithmic domain restrictions

Key concept: The function $f(x)=x^3+x^2-2x-1+\ln\left(1-\frac{x^2}{20}\right)$ combines polynomial growth with a logarithmic domain restriction.

Step by step

Domain first: the logarithm controls everything

For $f(x)=x^3+x^2-2x-1+\ln\left(1-\frac{x^2}{20}\right)$, the maximal domain comes from the logarithm. The polynomial part is defined for every real $x$, but the logarithm requires its inside to be positive:

$1-\frac{x^2}{20}>0.$

Solve that inequality:

$\frac{x^2}{20}<1 \quad \Longrightarrow \quad x^2<20 \quad \Longrightarrow \quad -\sqrt{20}<x<\sqrt{20}.$

So the maximal domain is

$(-\sqrt{20},\sqrt{20}).$

That interval is essential for the sketch because the graph cannot touch or cross the vertical boundaries at $x=\pm\sqrt{20}$.

Find the y-intercept exactly

The y-intercept happens when $x=0$. Substitute directly into the function:

$f(0)=0^3+0^2-2(0)-1+\ln\left(1-\frac{0^2}{20}\right).$

This simplifies to

$f(0)=-1+\ln(1).$

Since $\ln(1)=0$, we get

$f(0)=-1.$

So the y-intercept is $(0,-1)$. On the sketch, that point should be labeled clearly because it is the one exact coordinate you can locate without graphing software.

How to sketch the overall shape

The graph is a polynomial curve modified by a logarithmic term. Near the endpoints of the domain, the logarithm becomes very negative because its input approaches $0^+$, so the curve drops sharply as $x\to \pm\sqrt{20}$ from inside the domain. That creates two vertical asymptote-like barriers at the boundary points of the domain.

Inside the interval, the cubic term $x^3$ gives the graph a rising-right, falling-left tendency, while the quadratic and linear terms shift the shape. The logarithmic term bends the curve downward near the edges and adds a smooth dip where the input of the log is largest.

A useful way to sketch is to plot the intercept $(0,-1)$, mark the domain endpoints $\pm\sqrt{20}$ as excluded, and then estimate a few extra points symmetrically around zero. Because the logarithmic part depends on $x^2$, it is even, while the polynomial part is not symmetric. That means the full graph will not be symmetric about the y-axis.

Key features to label on the graph

You should label the domain restriction, the y-intercept, and the steep drop near the boundaries. If your instructor expects a fuller sketch, you can also estimate turning behavior numerically. But for a hand sketch, the strongest information is the domain and the intercept.

The crucial idea is that the logarithm never allows zero or negative inputs. Once you identify that restriction, the rest of the graphing task becomes much easier because the domain already tells you where the curve can and cannot exist.

Common graphing pitfall with this function

Many students start by expanding the polynomial and ignoring the logarithm until the end. That leads to the wrong sketch, because the logarithmic term is what limits the domain and forces the curve to end at finite x-values. Another mistake is forgetting that $\ln(1)=0$, which makes the y-intercept easier to find than it first appears.

If you keep the domain, intercept, and boundary behavior in view, the sketch will be much more accurate than a guess based on the polynomial alone.

Pitfall alert

The easiest place to make a mistake with $\ln\left(1-\frac{x^2}{20}\right)$ is the domain inequality. If you accidentally allow the logarithm’s input to be zero or negative, the sketch becomes invalid immediately. Another common error is reading the boundary points as actual intercepts, but $x=\pm\sqrt{20}$ are not included in the domain. Students also sometimes forget that the polynomial part does not remove the logarithmic restriction; the log still dominates the domain. When finding the y-intercept, do not overcomplicate $\ln(1)$: it is zero, so the intercept is exactly $(0,-1)$. For the sketch, avoid drawing the curve as if it were a simple cubic. The graph must stop inside the open interval and bend sharply downward near both endpoints because of the logarithmic term.

Try different conditions

If the function were changed to $g(x)=x^3+x^2-2x-1+\ln\left(1-\frac{x^2}{16}\right)$, the sketching process would stay the same, but the domain would change to $(-4,4)$ because $1-\frac{x^2}{16}>0$ requires $x^2<16$. The y-intercept would still be $(0,-1)$ since $\ln(1)=0$ at $x=0$. That modified problem shows why the denominator inside the logarithm matters so much: changing 20 to 16 moves the vertical boundary points closer to the origin and makes the graph end earlier. If the sign changed to $\ln\left(1+\frac{x^2}{20}\right)$, the domain would become all real numbers, which would produce a completely different graphing behavior.

Further reading

logarithmic domain restriction, y-intercept calculation, function sketching