Question

Hello
Considering positive values of x , locate the first two turning points.

(a) y= 1-sinx
(b) y=(1\2)x+ cos x

I am concerned about the 2nd point on each equation / How to find it out ?

Answer 1

i am not sure what "turning point" means here and why there should be two of them

Answer 2

rather than in infinite number

Answer 3

\[1-\sin(x)\] has a minimum value of \(0\) when \(\sin(x)=1\) i.e. when \(x=\frac{\pi}{2}\)

Answer 4

oops lost @Lama97

Answer 5

nop Iam here :) .. Turning points means stationary points :) well I got whats the min is but my question is asking for a max as well

Answer 6

@satellite73 I hope you're still around :)

Answer 7

yes the maximum value of \(1-\sin(x)\) is \(2\) when \(\sin(x)=-1\) i.e. when \(x=\frac{3\pi}{2}\)

Answer 8

clear? not sure if this was supposed to be a calculus question but it certainly is not one
the next may be though

Answer 9

But when using differentiation and equating the differentiated eqn to zero I get only one point

Answer 10

nop you're right :) its a calculus questin

Answer 11

@satellite73 You're right my friend :)

Answer 12

ok let me rant for a second
there is no point in taking the derivative
you get - cosine
big deal
you already know all about sine, so you do not need cosine
but if you want to set cosine equal to zero and solve, go ahead, you still get \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\)

Answer 13

how come I just got pie over two and not 3\2 pie :|

Answer 14

cause you didn't look hard enough i guess
cosine is zero two places on the interval \([0,2\pi)\)

Answer 15

got it ":| Thanks for clearing it up

Answer 16

this question required no calculus, but
\[f(x)=\frac{1}{2}x+\cos(x)\] will

Answer 17

derivative i think you got already as
\[f'(x)=\frac{1}{2}-\sin(x)\] if you set it equal to zero you solve
\[\sin(x)=\frac{1}{2}\] which occurs in more than on place in the interval \([0,2\pi)\)

Answer 18

clear right? once at \(\frac{\pi}{6}\) and also at \(\frac{5\pi}{6}\)

Answer 19

yeppp

Answer 20

Thanks so much . Appreciate your help :)