Posted by Yoona on Monday, January 2, 2012 at 1:55pm.
sorry for the double post
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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