Question

What are the absolute extrema of `f(x)=2cosx+sinx in[0,pi/2]`?

Answer 1

Absolute max is at `f(.4636) approx 2.2361`

Absolute min is at `f(pi/2)=1`

Explanation:

`f(x)=2cosx+sinx`

Find `f'(x)` by differentiating `f(x)`
`f'(x)=-2sinx+cosx`

Find any relative extrema by setting `f'(x)` equal to `0`:
`0=-2sinx+cosx`

`2sinx=cosx`

On the given interval, the only place that `f'(x)` changes sign (using a calculator) is at

`x=.4636476`

Now test the `x` values by plugging them into `f(x)`, and don't forget to include the bounds `x=0` and `x=pi/2`

`f(0) = 2`

`color(blue)(f(.4636) approx 2.236068)`

`color(red)(f(pi/2) = 1)`

Therefore, the absolute maximum of `f(x)` for `x in [0,pi/2]` is at `color(blue)(f(.4636) approx 2.2361)`, and the absolute minimum of `f(x)` on the interval is at `color(red)(f(pi/2)=1)`