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
Hello, camherokid!
Not that easy . . .
Note that x and tan(x) have the same sign (on that interval).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.
. . 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.