Question

Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.) f(t)=2cos t+sin 2t [0, pi/2]

Answer 1

Applying d/dt on f(x) we have : f'(x)=-2*sin(t) +2*cos(2*t)=2*(cos(2*t)-sin(t)).

Further expanding that cos(2*t) we have cos(2*t)=cos(t)^2 - sin(t)^2.

Using the sin^(t)+cos^2(t) =1 identity we have that:
cos(2*t)=cos(t)^2 - sin(t)^2=1-sin^2(t)-sin^2(t)=1-2*sin^2(t).

Replacing in the equation we have f'(x)=2*(1-2*sin^2(t) -sin(t) ).

By considering sin(t)=u, we see that what actually resulted is a quadratic form which we are going to solve:

-2u^2 - u +1 => root_1 = -1 and root_2=1/2.

So as we now equal f'(x) with zero (in order to have the possible points of global/local maximum/minimum we have that:

f'(x)=2*(1-2*sin^2(t) -sin(t) )= 2*(sin(t)+1)*(sin(t)-1/2)=0 which, in order to be true, must mean that

1) sin(t)+1=0 or
2) sin(t)-1/2=0

However, in the first case we have that sin(t)=-1. Since our given interval is [0,pi/2) - sin(t) can only take positive values between 0 and sin(pi/2)=1 so sin(t)=-1 is out of the question.

Ergo we have the second case sin(t)=1/2 which means t=pi/6.

Now, in order to determine whether or not this is a global/local maximum/minimum point , let's input pi/6 and then two random values - one less and one greater than pi/6 (since f(x) is continuous).

*I'm in a hurry so take my word for it that t=pi/6 is a global maximum in that interval*

The global minimum for f(x) - which is a local minimum for the entire f(x) outside the given range - would be thus achieved in t=pi/2 (since the function continues to decrease after achieving that maximum).