Problem: Sketch the graph of the function

Solution: We will use the procedure as outlined in Overview of graphing in Methods Survey - Graphing.

Step 1. The domain of this function is given by two conditions. First, the logarithm requires that x > 0; second, the fraction needs ln(x) not being zero, which means that we cannot have x = 1. From this we get the domain.

D( f ) = (0,1) ∪ (1,∞).

The function is continuous there. Since this set is not symmetric about the origin, the given function cannot be symmetric.

Intercepts: f (x) = 0 only if the numerator is zero, but x = 0 is not possible because of the logarithm in the denominator. Thus there is no x-intercept and also no y-intercept as 0 is not in the domain.

Step 2. We find limits at endpoints of the interval given by the domain.

Asymptotes: There are two candidates for vertical asymptotes, the proper endpoints x = 0 and x = 1. Since the one-sided limit at 0 converges, there is no vertical asymptote there. At 1 there is a one-sided limit that is improper (even both of them, but that is not needed), so there is a vertical asymptote at x = 1.

Because the limit at infinity diverges, there is no horizontal asymptote there, but the limits exist and therefore there is a chance for an oblique asymptote. We will use the appropriate algorithm.

The limit for A converges, so there was still a chance, but the limit for B diverges and therefore there is no oblique asymptote at infinity. By the way, we could have reached this conclusion from the first limit, without calculating B; the only A that could work is A = 0, but that would mean a horizontal asymptote and we already ruled this one out.

Step 3. We find the derivative and use it to determine monotonicity and local extrema.

Critical points: There are no points in the domain where the derivative does not exist; the derivative is zero when ln(x) = 1, that is, at x = e. The two intervals of the domain thus split into three intervals of monotonicity, we determine it using a table; we put closed endpoints when the function is continuous.

We have adjacent intervals of equal monotonicity, but since the connecting point 1 is in fact a hole in the domain, we cannot connect them into an interval regardless of how f goes. The conclusion is that f is decreasing on (0,1) and on (1,e] and increasing on [e,∞).

Local extrema: The given function has a local minimum f (e) = e.

Step 4. We find the second derivative and use it to determine concavity.

Dividing points: There are no points in the domain where the second derivative does not exist; the second derivative is zero when ln(x) = 2, that is, at x = e². The two intervals of the domain thus split into three intervals of concavity, we determine it using a table; we put closed endpoints when the function is continuous.

The conclusion is that f is concave down on (0,1) and on [e²,∞); it is concave up on (1,e²].

Inflection point: f (e²) = e²/2.

Step 5. Now we put it all together. First we put all points and limit trends that we obtained above into a picture. This will be the skeleton on which we will hang the function.

To see the shape of the graph better we combine the two tables above.

To make the picture more faithful we can find the limit of the derivative at 0 from the right.

This tells us that as the graph starts off the origin, it should hug the x-axis. Now we are ready to sketch the graph.

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