calculus

2. Consider the function f defined by f(x)=(e^X)cosx with domain[0,2pie] .
a. Find the absolute maximum and minimum values of f(x)
b. Find the intervals on which f is increasing.
c. Find the x-coordinate of each point of inflection of the graph of f.

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  1. sorry for the double post

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  2. f(x) = e^x (cosx)
    f'(x) = e^x(-sinx) + e^x(cosx)
    = e^x(cosx - sinx) = 0 for max/min values of f(x)
    so e^x = 0 , no solution
    or
    cosx - sinx = 0
    sinx= cosx
    sinx/cosx = 1
    tanx = 1
    x = π/4 or 5π/4

    f(0) = e^0(cos0) = 1
    f(2π) = e^(2π)(1) = e^(2π) = appr. 535.5
    f(π/4) = e^(π/4) (√2/2) = appr. 1.55
    f(5π/4) = e^(5π/4) cos 5π/4 = appr. -35.9

    take it from there

    b) the function is increasing when f'(x) > 0
    e^x(cosx - sinx) > 0
    since e^x > 0 for all x
    this results in cosx - sinx >0
    -sinx > -cosx
    sinx/cosx < 1
    tanx < 1
    So for the domain from 0 to 2π
    tanx < 1 for
    0 < x < π/4 OR π/2 < x < 5π/4 OR 3π/2 < x < 2π
    ( I looked at the tangent curve for these)

    c) take the derivative of f'(x)
    f''(x) = e^x(-cosx) + e^x(-sinx) + e^x(-sinx) + e^x(cosx)
    = -2e^x sinx
    = 0 for pts of inflection

    then sinx = 0
    x = 0 , π, 2π

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