posted by fareha on .
4. Given the function f defined by f(x) = cos2x for -π≤ x ≤π
a. Find the x-intercepts of the graph of f.
b. Find the x and y coordinates of all relative maximum points of f. Justify your answer.
c. Find the intervals on which the graph of f is increasing.
Make a rough sketch of y = cos 2x
x-intercepts, when cos 2x = 0
2x = π/2 or 2x = 3π/2
x = π/4 or x = 3π/4
period of cos2x is π, thus adding/subracting π to any answer will yield a new answer.
π/4 - π = -3π/4
3π/4 - π = -π/4
x-intercepts are ±π/4, ±3π/4
f'(x) = 2sin 2x
= 0 for max/min
2sin 2x = 0
sin 2x= 0
2x = 0 , π , 2π
x = 0, π/2, π , again, since the period is π other answers are
x = -π/2, 0, -π
A quick look at your y = cos 2x will show you which of these are maximums and which are minimum
Max: at x = -π, 0, π
max points: (-π,2), (0,2), and (π,2)
To show where y = cos 2x is increasing,
f'(x) has to be positive
f'(x) = -2sin 2x has to be positive
so take a look where y = -2sin 2x lies above the x-axis
Since you have the sketch of y = cos 2x, it is easy to see that the curve is increasing between -π/2 and 0 and again between π/2 and π