I\'m stuck on finding the intercepts, please help me with this problem: Consider
ID: 2829219 • Letter: I
Question
I'm stuck on finding the intercepts, please help me with this problem:
Consider the function f(x)=3cosx?cos3x for 0<x<2?.
For the following questions, write inf for ?, -inf for ??, U for the union symbol, None if no answer exists, and separate by a comma if more than one answer exist.
a.) The x-intercepts are .
b.) f?(x)= .
c.) f(x) is increasing on the interval(s) .
d.) f(x) is decreasing on the interval(s) .
e.) f(x) has a local maximum at x = .
f.) f(x) has a local minimum at x = .
g.) f??(x)= .
h.) f(x) is concave up on the interval(s) .
i.) f(x) is concave down on the interval(s) .
j.) The x-coordinate of the points of inflection are .
Explanation / Answer
x intercepts are f(x) = 0 at, x = pi/2 and 3pi/2
f(x) = 6cosx - 4cos^3(x)
f''(x) = -6sinx + 12 cos^2(x)sinx = -6sinx cos2x positive for pi/4 to 3pi/4 and pi to 5pi/4 and 7pi/4 to 2 pi
increasing on these intervals and decreasing on rest
local maximum at 3pi/4 ,5pi/4, 2 pi
local minimum at pi/4, pi, 7 pi/4
f''(x) = -6cosx cos2x + 12 sinx sin2x = -6cos3x + 6sinx sin 2x
concave up on interval where f'(x) greater than 0 , concave down for f''(x) less than 0 and point of inflection at the point where greater than zero transforms to less than or vice versa.