Solving task on-line

There is a way of constructing graph of a function based on the analytical study of the function. Research conducted by the following approximate scheme:
1) elucidation of the function definition;
2) addressed the issue of even or odd functions;
3) investigate the frequency of functions;
4) find the point of intersection of the curve with the coordinate axes;
5) find the point of discontinuity and determine their character;
6) conduct research on an extreme, are the extreme values of functions;
7) look for the point of inflection and intervals of convexity and concavity of the curve;
8) finding the asymptotes of the curve;
9) results are applied to the drawing and receive a schedule of the studied function.

Sample. Conduct a complete study of the function  and construct its graph.

1) The function is defined everywhere except at points .

2) The function is odd, since f(-x) = -f (x), and, hence, its graph is symmetric about the origin. Therefore confine ourselves to study only for 0 ≤ x ≤ +∞.

3) Function is not periodic.

4) Since y = 0 only when x = 0, then the intersection with the coordinate axes occurs only at the origin.

5) The function has a discontinuity of the second kind at , with , . In passing, we note that the direct  – vertical asymptote.

6) Находим  and equate it to zero: , where x₁ = -3, x₂ = 0, x₃ = 3. At the extreme need to investigate only the point x = 3 (point x₂ = 0 does not investigate, because it is a boundary point gap [0, +∞)).

In the neighborhood of point x₃=3 has: y’>0 при x<3 and y ’<0 at x>3, hence the point x₃ function has a maximum, y_max(3)=-9/2.

Find the first derivative of the function

To verify finding the minimum and maximum values.

7) Find . We see that y’’=0 only at x = 0, while y”<0 at x<0 and y”>0 at x>0, hence, the point (0,0) curve has an inflection. Sometimes the direction of concavity may change when passing through the gap curve, so you should find the sign of y” and some points of discontinuity. In our case y”>0 on the interval  and y”<0 on , hence, on  curve is concave and convex on .

Найти вторую производную функции

8) Elucidate the question of the asymptote.

The presence of vertical asymptotes  finding above. Looking horizontal: , hence, no horizontal asymptotes.

Find the oblique asymptote: , , hence, y=-x – Bilateral oblique asymptote.

9) Now, using the data obtained, we construct a drawing:

Construct a graph of the function