3 years ago

Find the extrema of the function y=cos2x

Mathematics

1 answer:

marishachu [46]3 years ago

4 0

Answer:

Step-by-step explanation:

Find derivative of the function:

y' = -2·sin(2·x)

-2·sin(2·x) = 0

x1 = 0

x2 = 3.142

f(0) = 1

f(3.142) = 1

Answer

fmin = 1, fmax = 1

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Show that if x and y may both be written as the sum of the squares of two rational integers, then their product xy may also be w

Georgia [21]

Answer:

See proof below

Step-by-step explanation:

One way to solve this problem is to "add a zero" to complete the required squares in the expression of xy.

Let $x=m^2+n^2$ and $y=l^2+k^2$ with $m,n,l,k\in \mathbb{Z}$ . Multiplying the two equations with the distributive law and reordering the result with the commutative law, we get $xy=(m^2+n^2)(l^2+k^2)=m^2l^2+m^2k^2+n^2l^2+n^2k^2=n^2l^2+m^2k^2+m^2l^2+n^2k^2$

Now, note that $0=2lkmn-2lkmn=2nkml-2nlmk$ by the commutativity of rational integers. Add this convenient zero the the previous equation to obtain $xy=n^2l^2-2nlmk+m^2k^2+m^2l^2+2nkml+n^2k^2=(nl-mk)^2+(ml+nk)^2$ , thus xy is the sum of the squares of $nl-mk,ml+nk\in \mathbb{Z}$ .