Question

I need to find Absolute max and min
f(x) = sin2x + 2cosx on [0, π/2]

Answer 1

Is that \(sin^{2}(x)\;or\;\sin(2x)\)?

Have you considered a 1st Derivative? You WILL have to check the endpoints.

Answer 2

sin(2x)

Answer 3

i got cos2x + 2sinx for derivative

Answer 4

See how much more clear it is WITH the parentheses? Now, about that derivative...

s/b \(f'(x) = 2\cos(2x) - 2\sin(x)\)

Something went wrong with both pieces.

Answer 5

how u got 2cos(2x)?

Answer 6

It's the Chain Rule.

\(\dfrac{d}{dx}\sin(2x) = \cos(2x)\cdot \dfrac{d}{dx}(2x)\)

Answer 7

oh thanks

Answer 8

how do i get the max and min

Answer 9

do i out 2cos(2x) = 0 and -2sinx = 0 ?

Answer 10

put*

Answer 11

No, that works ONLY with FACTORS, not terms.

If it were me, I would convert \(\cos(2x)\) to \(1−2\sin^{2}(x)\) and treat it like a quadratic equation.

Answer 12

can you please show me all the way? I dont understand much