You have been asked to determine whether the function f(x)= 3 + 4cosx +cos2x is ever negative.
a) Explain why you need to consider values of x only in the interval [0,2π ].
b) Is f ever negative? Explain.
This is extra credit and I can't seem to figure it out. I'm used to find the derivative of polynomials.
I hope you have come across the identity
cos2x = 2 cos^2 x - 1
so
3 + 4cosx +cos2x
= 3 + 4cosx + 2cos^2x - 1
= 2cos^2x + 4cosx + 2
= 2(cos^2x + 2cosx + 1)
= 2(cosx+1)^2
something squared >0
so 2(something squared)>0
and f(x) can thus never be negative
the above answers part b)
for a) the period of cosx is 2π and the period of cos2x is π
so the period of the whole function is 2π
So whatever happens will happen between 0 and 2π, and simply repeat itself after that.
what do you mean by something squared?
To determine whether the function f(x) = 3 + 4cos(x) + cos(2x) is ever negative, we can follow these steps:
a) Explain why you need to consider values of x only in the interval [0, 2π ].
The reason we consider values of x only in the interval [0, 2π] is because the cosine function has a period of 2π. This means that the values of cosine repeat after each interval of 2π. Therefore, analyzing the function within one full period is sufficient to evaluate its behavior.
b) Is f ever negative? Explain.
To determine if f(x) is ever negative, we need to find the values of x in the interval [0, 2π] for which f(x) < 0. To do this, we can analyze the behavior of the cosine function.
The range of the cosine function is [-1, 1]. This means that the maximum value of 4cos(x) will be 4, and the minimum value is -4. Similarly, the maximum value of cos(2x) will be 1, and the minimum value is -1.
Adding these two terms to the constant 3, we have:
f(x) = 3 + 4cos(x) + cos(2x)
Now, to find the minimum value of f(x), we want to find the smallest possible sum of these terms. The minimum occurs when both cos(x) and cos(2x) are at their minimum values (-1 and -1, respectively).
So, the minimum value of f(x) is:
f(x) = 3 + 4(-1) + (-1) = 3 - 4 - 1 = -2
Since -2 is greater than 0, we conclude that the minimum value of f(x) is greater than or equal to 0. Therefore, f(x) is never negative in the interval [0, 2π].
Hence, the function f(x) = 3 + 4cos(x) + cos(2x) is never negative for x in the interval [0, 2π].
To determine whether the function f(x) = 3 + 4cos(x) + cos(2x) is ever negative, we need to consider the following steps:
a) Explain why you need to consider values of x only in the interval [0,2π ]:
In trigonometry, the cosine function has a periodic nature, which means its values repeat after certain intervals. The cosine function oscillates between the values of -1 and 1. By considering values of x only in the interval [0, 2π], we cover one complete period of the cosine function, ensuring that we analyze all possible values of the cosine function within this range. Additionally, it simplifies our analysis and allows us to examine the function over a manageable interval, rather than considering an infinite range.
b) Is f ever negative? Explain:
To determine if f(x) is ever negative, we can evaluate the function at different points within the interval [0, 2π] and check if any of these evaluations result in a negative value. By observing the function's behavior, we can infer its overall behavior throughout the interval.
Let's evaluate the function at various points:
For x = 0,
f(0) = 3 + 4cos(0) + cos(2*0) = 3 + 4(1) + 1 = 8
For x = π/2,
f(π/2) = 3 + 4cos(π/2) + cos(2*(π/2)) = 3 + 0 + cos(π) = 2
For x = π,
f(π) = 3 + 4cos(π) + cos(2*π) = 3 + 4(-1) + 1 = -2
For x = 3π/2,
f(3π/2) = 3 + 4cos(3π/2) + cos(2*(3π/2)) = 3 + 0 + cos(3π) = 2
For x = 2π,
f(2π) = 3 + 4cos(2π) + cos(2*(2π)) = 3 + 4(1) + 1 = 8
Analyzing these evaluations, we observe that f(x) takes on both positive and negative values within the interval [0, 2π]. Specifically, at x = π, the function f(x) equals -2, which means it reaches a negative value within the given interval.
Thus, the function f(x) = 3 + 4cos(x) + cos(2x) is indeed negative at x = π and is therefore negative at some point within the interval [0, 2π].