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
or
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

a. To find the x-intercepts of the graph of f, we need to find the values of x for which f(x) = cos^2(x) equals zero.

When cos^2(x) = 0, it means that cos(x) = 0, since squaring zero still gives zero.

The values of x for which cos(x) = 0 are π/2 and 3π/2, since cos(x) equals zero at these points on the unit circle.

Therefore, the x-intercepts of the graph of f are π/2 and 3π/2.

b. To find the relative maximum points of f, we need to find the local maximum points on the graph.

Since f(x) = cos^2(x), the maximum value of cos^2(x) is 1, while the minimum value is 0.

The graph of f(x) = cos^2(x) is a repetitive curve that oscillates between 0 and 1.

This implies that there are no relative maximum points, as there is no point on the graph that is higher than all its neighbors.

c. To find the intervals on which the graph of f is increasing, we need to examine the derivative of f(x) = cos^2(x).

Taking the derivative of cos^2(x) gives us:

f'(x) = -2sin(x)cos(x).

The graph of f'(x) = -2sin(x)cos(x) oscillates between positive and negative values, indicating that f(x) is both increasing and decreasing over different intervals.

Therefore, we can say that the graph of f is increasing on intervals where -2sin(x)cos(x) > 0. These intervals occur when either -π/2 < x < 0 or π/2 < x < π.

To find the x-intercepts of the graph of f, we need to solve the equation f(x) = 0.

a. Solving f(x) = cos^2(x) = 0:
cos^2(x) = 0
cos(x) = 0
x = ±π/2

So, the x-intercepts of the graph of f are x = -π/2 and x = π/2.

To find the relative maximum points of f, we need to find the critical points where the derivative of f equals zero or is undefined.

b. Finding the derivative of f(x):
f'(x) = -2sin(2x)

Setting f'(x) = 0 to find the critical points:
-2sin(2x) = 0
sin(2x) = 0
2x = kπ, where k is an integer
x = kπ/2, where k is an integer

Since the given range is -π ≤ x ≤ π, we only need to consider -π/2, 0, and π/2.

Evaluating the function f at these critical points:
f(-π/2) = cos^2(-π/2) = cos^2(-90°) = cos^2(-1.5708) = 0
f(0) = cos^2(0) = cos^2(0°) = cos^2(0) = 1
f(π/2) = cos^2(π/2) = cos^2(90°) = cos^2(1.5708) = 0

The x and y coordinates of the relative maximum points are:
(-π/2, 0) and (π/2, 0).

The function f(x) = cos^2(x) has relative maximum points at x = -π/2 and x = π/2, with a y-coordinate of 0. This is because at these points, the function reaches its highest value (since the cosine function is bounded between -1 and 1, its square will always be non-negative).

To find the intervals on which the graph of f is increasing, we need to find the intervals where the derivative f'(x) is positive.

c. Analyzing the sign of f'(x) on the interval -π ≤ x ≤ π:
Since f'(x) = -2sin(2x), the sign of f'(x) will depend on the sign of sin(2x).

sin(2x) is positive when 0 < x < π/2 and π < x < 3π/2.
sin(2x) is negative when -π/2 < x < 0 and π/2 < x < π.

Thus, the graph of f is increasing on the intervals (0, π/2) and (π, 3π/2).

a. To find the x-intercepts of the graph of f, we need to solve the equation f(x) = 0. In this case, f(x) = cos^2(x).

We know that cos^2(x) = 0 when cos(x) = 0, because the square of any number is 0 if and only if the number itself is 0.

For cos(x) = 0, we need to find the values of x within the given interval where the cosine function is equal to 0. In the interval -π ≤ x ≤ π, the cosine function will be equal to 0 at x = -π/2 and x = π/2.

Therefore, the x-intercepts of the graph of f are -π/2 and π/2.

b. To find the relative maximum points of f, we need to analyze the critical points of the function. In this case, the critical points are the values of x where the derivative of f(x) changes sign.

Let's start by finding the derivative of f(x):

f(x) = cos^2(x)

Using the chain rule, we can find the derivative:

f'(x) = 2cos(x)(-sin(x))

Simplifying:

f'(x) = -2sin(x)cos(x)

Now, let's find the critical points by setting the derivative equal to 0 and solving for x:

-2sin(x)cos(x) = 0

This equation is satisfied when either sin(x) = 0 or cos(x) = 0.

For sin(x) = 0, the solutions within the given interval are x = 0 and x = π.

For cos(x) = 0, the solutions within the given interval are x = -π/2 and x = π/2.

Now, we need to determine which of these critical points are relative maximum points by analyzing the behavior of the function around these points. We can use the second derivative test for this.

The second derivative of f(x) can be found by taking the derivative of f'(x):

f''(x) = (-2cos(x)cos(x)) + (-2sin(x)(-sin(x))) = -2cos^2(x) + 2sin^2(x)

Simplifying further:

f''(x) = -2cos^2(x) + 2(1-cos^2(x)) = -4cos^2(x) + 2

Now, let's evaluate the second derivative at each critical point.

At x = 0: f''(0) = -4cos^2(0) + 2 = -4(1) + 2 = -2 < 0. This means that at x = 0, the function has a relative maximum.

At x = π: f''(π) = -4cos^2(π) + 2 = -4(-1) + 2 = 6 > 0. This means that at x = π, the function has a relative minimum.

At x = -π/2: f''(-π/2) = -4cos^2(-π/2) + 2 = -4(0) + 2 = 2 > 0. This means that at x = -π/2, the function has a relative minimum.

At x = π/2: f''(π/2) = -4cos^2(π/2) + 2 = -4(0) + 2 = 2 > 0. This means that at x = π/2, the function has a relative minimum.

Therefore, the only relative maximum point is at x = 0.

The x-coordinate of the relative maximum point is 0, and the corresponding y-coordinate can be found by evaluating the function at x = 0:

f(0) = cos^2(0) = 1^2 = 1.

So the relative maximum point is (0, 1).

c. To find the intervals on which the graph of f is increasing, we need to determine where the derivative of f(x) is positive.

From part b, we found that the derivative of f(x) is f'(x) = -2sin(x)cos(x).

To determine where f'(x) > 0, we need to find the intervals where sin(x) < 0 and cos(x) > 0. In other words, we are interested in the values of x where the sine function is negative and the cosine function is positive.

In the given interval -π ≤ x ≤ π, the sine function is negative in the interval -π/2 < x < 0 and the cosine function is positive in the intervals 0 < x < π/2 and π/2 < x < π.

Therefore, the graph of f is increasing in the intervals -π/2 < x < 0, 0 < x < π/2, and π/2 < x < π.