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. Since arctangent accepts any argument, the only problem is with the fraction. Thus the domain is

D( f ) = (−∞,0) ∪ (0,∞)

and the function is continuous there. Since the only place where the variable comes in is an even power, this function is even. Indeed, it is easy to show that f (−x) = f (x).

The x-intercept: The arctangent is zero only if its argument is zero, which is not possible. Thus there is no x-intercept, and since 0 is not in the domain, there is also no y-intercept.

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

Actually, since the function is even, it was enough to calculate only the first two limits and the other two follow by symmetry of the function. We see that we could actually "fill in" the hole in the graph, that is, if we also defined f (0) = π/2, we would get a continuous function on the whole real line.

Asymptotes: Since the function is continuous on its domain, the only candidate for a vertical asymptote is the proper endpoint x = 0. However, we have a proper limit there, so there is no vertical asymptote.

Because limits at negative infinity and infinity converge, there are horizontal asymptotes there, and from the results of these limits we conclude that the line y = 0 is the horizontal asymptote at both negative and positive infinity.

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 at x = 0, but that point is not in the domain, so it is not a critical point after all. There are no critical points, so the two intervals from the domain are also intervals of monotonicity. We determine it using a table.

The conclusion is that f is increasing on (−∞,0) and decreasing on (0,∞).

Local extrema: There is no local extreme (monotonicity changes at 0, but that point is not in the domain).

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 at x = −1 and at x = 1. The domain thus splits into four 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 up on (−∞,−1] and on [1,∞); it is concave down on [−1,0) and on (0,1].

Inflection points: There are two, at −1 and at 1. The values are

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.

The sketch will be more faithful if we also find out what are the one-sided limits of the derivative at 0 and derivatives at −1, 1.

We see that the graph curves toward that hole at 0 as if there was a horizontal tangent line there (if we extended the definition to a continuous function on the real line as discussed above, there would be a horizontal tangent line at 0, also a local maximum). We also see that the graph goes through the points of inflection a bit less steeply than at 45 degrees. Now we are ready to sketch the graph.

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