Posted by **Yoona** on Monday, January 2, 2012 at 1:55pm.

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.

- calculus -
**Yoona**, Monday, January 2, 2012 at 1:56pm
sorry for the double post

- calculus -
**Reiny**, Monday, January 2, 2012 at 5:19pm
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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