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Math Help - Find Vertical and Horizontal Asymtot

  1. #1
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    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
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  2. #2
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    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.

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