what are the intercepts of the graph of f(x)=4cos^2 -3 on the interval [0, 2pi]?
I am sure you meant:
f(x)=4cos^2 x -3
for the x-intercepts:
4cos^2 x -3 = 0
cos^2 x = 3/4
cos x = ±√3/2 ---> so x is in all 4 quadrants.
I know that cos π/6 = √3/2
so x = π/6, π - π/6, π + π/6, and 2π-π/6
or
x = π/6, 5π/6, 7π/6, and 11π/6
for the y-intercept, let x = 0
f(0) = 4cos^2 0 - 3
= 4(1) - 3 = 1
To find the intercepts of the graph of the function f(x) = 4cos^2(x) - 3 on the interval [0, 2π], we need to determine the values of x where the graph intersects or crosses the x-axis.
The x-intercepts occur when the y-value (or the function value) is equal to zero. So, we need to solve the equation f(x) = 0.
The given function is f(x) = 4cos^2(x) - 3. To find the x-intercepts, set f(x) equal to zero:
0 = 4cos^2(x) - 3.
Now, to solve this equation, we can isolate the cosine term and solve for cos(x). We start by adding 3 to both sides of the equation:
3 = 4cos^2(x).
Next, divide both sides by 4:
3/4 = cos^2(x).
To solve for cos(x), take the square root of both sides of the equation:
±sqrt(3/4) = cos(x).
Now, let's simplify the right side by taking the positive square root and the negative square root individually:
cos(x) = ±sqrt(3)/2.
The values of cos(x) that satisfy this equation are ±sqrt(3)/2. We can find the corresponding values of x by taking the inverse cosine (also called arccosine) of these values. Using a calculator or a reference table, we find that the inverse cosine of sqrt(3)/2 is π/6 and the inverse cosine of -sqrt(3)/2 is 5π/6.
Therefore, the x-intercepts on the interval [0, 2π] occur at x = π/6 and x = 5π/6.
So, the intercepts of the graph of f(x) = 4cos^2(x) - 3 on the interval [0, 2π] are x = π/6 and x = 5π/6.