"Maximum and minimum of f(x)=2sin2x+2cos4x" was answered by students like you

Lucille Davidson

Answered question

2022-01-16

Maximum and minimum of

f (x) = 2 \sin^{2} x + 2 \cos^{4} x

Bertha Jordan

Beginner2022-01-17Added 37 answers

You also could linearise the function first, doing some trigonometry:

2 \sin^{2} x + 2 \cos^{4} x = 1 - \cos 2 x + {\frac{12}{2 \cos^{2} x}}^{2} =

= 1 - \cos 2 x + \frac{12}{+} \cos 2 x + {\frac{12}{\cos}}^{2} 2 x

= \frac{32}{+} \frac{1 + \cos 4 x}{4} = \frac{7 + \cos 4 x}{4}

Now it is obvious this expression has a global minimum when

\cos 4 x = - 1

and a global maximum when

\cos 4 x = 1

RizerMix

Expert2022-01-20Added 656 answers

f (x) = 2 \sin^{2} + 2 (1 - \sin^{2}) \cos^{2} x =

2 (\sin^{2} x + \cos^{2} x) - 2 \sin^{2} x \cos^{2};

f (x) = 2 - (\frac{1}{2}) \sin^{2} (2 x);

Minimum: at

2 x = k 2 π + \frac{π}{2} k \in Z

;

f_{m i n} (x) = \frac{3}{2};

Maximum: at

2 x = k 2 π, k \in Z;

f_{m a x} (x) = 2

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