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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Strike441 [17]

Answer:

Step-by-step explanation:

These problems are based on triangle ratios. You cannot use the Pythagorean theorem to solve them.

The first triangle is a 45 45 90 degree triangle (I'm talking about the angles), and so, the ratio is 1:1: $\sqrt{2\\}$ , so I have to divide the hypotenuse by $\sqrt{2\\}$ to get the legs. The hypotenuse is 15 $\sqrt{6}$ , so that divided by $\sqrt{2\\}$ is 15 $\sqrt{3\\}$ . X is the same length as y because of the triangle ratio, so both x and y for the first triangle are 15 $\sqrt{3\\}$ .

The second triangle is a 30 60 90 degree triangle, so the ratio is x:x $\sqrt{3}$ :2x. The short leg is 7 $\sqrt{3}$ , so 7 $\sqrt{3}$ * 2 is the hypotenuse, which is 14 $\sqrt{3}$ . The long leg is 7 $\sqrt{3}$ * $\sqrt{3}$ , which is 21. So, x for the second triangle is 14 $\sqrt{3}$ , and y for the second triangle is 21.