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 the denominator in the fraction is always at least 1, the only condition on x is imposed by the arcsine, it requires that its argument is between −1 and 1. It turns out that this is always true, we already looked at this function in this problem in Solved Problems - Derivative and concluded that the domain of f is the whole real line. We note that the function is continuous there.

What can we say about symmetry? Since arcsine is an odd function and we have a fraction with odd numerator and even denominator inside, an experienced function investigator would say that f is clearly odd. We will try it the safe way, we substitute -x into f and see what happens.

We just proved that f is an odd function.

The x-intercept: Arcsine is zero only if its argument is zero, and the fraction inside is zero only for x = 0. We see that f (0) = 0 is the only x-intercept and it is also the y-intercept.

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

Actually, it was not necessary to calculate the second limit, since it follows from the symmetry.

Asymptotes: Since the function is continuous on the whole real line, there cannot be vertical asymptotes. Because limits at negative infinity and infinity converge, there are horizontal asymptotes there. The limit results show that the line y = 0 is the horizontal asymptote of f at both negative and positive infinity.

Step 3. We find the derivative and use it to determine monotonicity and local extrema. In fact, finding derivative is surprisingly tricky, we did it as a problem of its own in Solved Problems - Derivative. We eventually concluded that

and that there is no derivative at −1 and at 1.

Critical points: The derivative does not exist at −1 and at 1; other critical points are those that make the derivative equal to zero, but there are no such points. Thus we have critical points x = −1 and x = 1 and the domain splits into three intervals of monotonicity. We determine monotonicity using a table; we put closed endpoints when the function is continuous.

We conclude that f is decreasing on (−∞,−1] and on [1,∞); it is increasing on [−1,1].

Local extrema: The given function has a local minimum f (−1) = −π/2 and a local maximum f (1) = π/2.

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

Dividing points: There are two points in the domain where the second derivative does not exist, namely x = −1 and x = 1. There is also a point where the second derivative is zero, x = 0. 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 down on (−∞,−1] and on [0,1]; it is concave up on [−1,0] and on [1,∞).

Inflection points: f (−1) = −π/2, f (0) = 0, and f (1) = π/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.

We get a more faithful sketch if we also use the knowledge of one-sided derivatives at −1 and 1.

We see that at −1 and at 1 the curves connect at right angles. Now we are ready to sketch the graph.

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