# Thread: Find Vertical and Horizontal Asymtot

1. ## Find Vertical and Horizontal Asymtot

a. Find Vert and Hori asymtot
b. Interval of increase and decrease
c. Local Max and Min
d. Concavity and Inflection point.

f(x)= x Tan[x], -Pi/2 < x < Pi/2

2. Hello, camherokid!

Not that easy . . .

f(x) .= .x·tan(x), . -π/2 < x < π/2

a. Find vertical and horizontal asymtotes.
b. Intervals of increase and decrease
c. Local Max and Min
d. Concavity and Inflection point.
Note that x and tan(x) have the same sign (on that interval).
. . Hence, f(x) is always positive (or zero).

The graph is a "U" with its vertex at the origin
. . and vertical asymptotes at x = -π/2 and x = π/2
. . which are not in the interval.

(a) There are no vertical or horizontal asymptotes.

The derivative is: .f'(x) .= .x·secē(x) + tan(x)

There is a horizontal tangent at x = 0.

We know that secē(x) is always positive
. . and that x and tan(x) have the same sign.
Hence: if x < 0, then f'(x) < 0
. .and: if x > 0, then f'(x) > 0

(b) The function is decreasing on (-π/2,0)
. . . . . . . . . . .and increasing on (0, π/2)

(c) There is an absolute minimum at (0,0).
. . .There is no maximum.

The second derivative is: .f''(x) .= .2x·secē(x)·tan(x) + 2·secē(x)

We have: .f''(x) .= .2·secē(x) [x·tan(x) + 1]

We know that secē(x) is always positive.
Since x and tan(x) have the same sign, x.tan(x) is positive (or zero).
. . Hence, x·tan(x) + 1 is always positive.
Therefore, f''(x) is always positive.

(d) The graph is always concave up.
. . .There are no inflection points.