Asked by John
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.
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.
Answers
Answered by
Reiny
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.
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.
Answered by
John
what do you mean by something squared?
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