Find the maximum and minimum values for the function $f$ defined by $f(x) = 2sinx + cos2x$ in the interval $[0, pi/2]$ ?

Question

36874 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T15:52:13

Minimum:
$f(x) = {1, x=0, pi/2}$
Maximum:
$f(x) = {3/2, x=pi/6}$

$f(x) = 2sinx + cos2x$

$f'(x) = 2cosx - 2sin2x$

Now to find the critical number,

$f'(x) = 0$

$2cosx - 2sin2x = 0$
$2cosx-2(2sinxcosx) = 0$
$2cosx - 4sinxcosx = 0$
$2cosx(1-2sinx) = 0$

$2cosx=0$
$x = pi/2$

and $1-2sinx=0$
$x=pi/6$

Therefore,

When $x=0$ , $f(x) = 2.0+1 = 1$ and
when $x=pi/2, f(x) = 2.1-1=1$

And

when $x=pi/6, f(x) = 2.1/2+1/2 = 3/2$

Hence,

Minimum:
$f(x) = {1, x=0, pi/2}$
Maximum:
$f(x) = {3/2, x=pi/6}$

Find the maximum and minimum values for the function ff defined by f(x) = 2sinx + cos2xf(x) = 2sinx + cos2x in the interval [0, pi/2][0, pi/2]?