Hello, camherokid!

Not that easy . . .

Note thatf(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.xandtan(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 arenotin 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.