Question

In: Math

f (x) = -0.248226*cos (2 x) - 0.0184829*cos ((2+2)x) - 0.0594608*cos(x)*sin(x) + 0.123626*sin ((2+2)x). The intervall...

f (x) = -0.248226*cos (2 x) - 0.0184829*cos ((2+2)x) - 0.0594608*cos(x)*sin(x) + 0.123626*sin ((2+2)x).

The intervall is ]0, 3/2[

What is the local maximum and local minimum? Answer with 5 decimals

Solutions

Expert Solution

We use 2nd derivative test to find local Maxima and minima.

f'(x)= d/dx[-0.248226*cos (2 x) - 0.0184829*cos (2+2x) - 0.0594608*cos(x)*sin(x) + 0.123626*sin ((2+2x)]

Here we write cos(x)*sin(x) = (1/2)sin(2x) = 0.5sin(2x)

Therefor, 0.0594608*cos(x)*sin(x) = 0.0297304 sin(2x)

Here we get,

milcah answered 2 years ago
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