Question

what is the largest value of the function f(x)=sin(sinx)?

Answer 1

got options ?

Answer 2

so this is what i did:
y=sin(sinx)
sinx=u(x)
u'(x)=cosx
y'=(sin(u))'*u'(x) = sinx*cosx
and sinx*cosx=sin2x/2
so i got that y'=sin2x/2 and this is where i'm stuck

Answer 3

the correct answer should be sin1

Answer 4

0 \(\le\) sin x\(\le\) 1
:D
take sin x for all
sin (0) \(\le\) sin (sin x)\(\le\) sin (1)

Answer 5

so this is what i did:
y=sin(sinx)
sinx=u(x)
u'(x)=cosx
y'=(sin(u))'*u'(x) = sinx*cosx ***************
and sinx*cosx=sin2x/2
so i got that y'=sin2x/2 and this is where i'm stuck

Answer 6

that like has a mistake

Answer 7

you should get :
y' = cos(sinx) * cosx

Answer 8

set it equal to 0 and find your critical points

Answer 9

so cos(sinx)*cosx=0
and is there a way to simplify this or?! :/

Answer 10

use zero product property :

ab = 0 means
1) a = 0
or
2) b = 0
or
3) a, b = 0

Answer 11

cos(sinx)*cosx=0 means

cos(sinx) = 0 OR cosx = 0

Answer 12

rightt so dor cos(sinx)=0 sinx=1

Answer 13

for*

Answer 14

i'm so slow today xD

Answer 15

nope, try again

Answer 16

\(\cos(\sin x) = 0\)

\(\sin x = \arccos (0) = \dfrac{\pi}{2}\)

\(x = \arcsin (\frac{\pi}{2}) = 1\)

Answer 17

solve the other case also :

cosx = 0

Answer 18

@rational is the domain /range method work ?

Answer 19

does*

Answer 20

it works right ?

Answer 21

but he is trying to solve it using calculus... so im just trying to help him solve it in his own way..

Answer 22

ic..

Answer 23

solving cosx = 0 gives x = pi/2
so the critical points are :
\(\large x = 1, \pi /2 \)

Answer 24

evaluate f(x) at these points and the maximum value u get is the maximum value of f(x)

Answer 25

f(x) = sin(sinx)
f(1) = ?
f(pi/2) = ?