Question

Find the maximum and minimum values of the function f(θ) = 5√2θ − 5 sec on the interval 0, π/3  (Round your answers to three decimal places.) maximum f(θ) =  minimum f(θ) =

Answer 1

Substitute x in the place of

Multiply each side by negative one

Now use quadratic formula

But the range of sinx is [ 1 , -1 ] we can have only

By the second derivative it is locally minimum at

Substitute in given f(x)

Now check the end points f(0) = -5 and

= -2.595

So the maximum value of  

= -1.517

And the minimum value is f(o) = -5