values of c

[Ryan Rigdon]

if i have the following with a given interval of [-?,?]

sin^(-1) (0) = c


Values of c = ?


I answered it with c = ?, 2?


this is where sin is 0 on the unit circle.

 

[Deleted member 4993]

if i have the following with a given interval of [-?,?]

sin^(-1) (0) = c


Values of c = ?


I answered it with c = ?, 2? <<< This number is not within the given domain


this is where sin is 0 on the unit circle.

 

[Ryan Rigdon]

oops my mistake. so i would have an answer of

No values of c in the given domain satisfy the Mean Value Theorem

 

[Deleted member 4993]

oops my mistake. so i would have an answer of

No values of c in the given domain satisfy the Mean Value Theorem

Now you are confusing me - where is MVT coming into this picture?

MVT involves derivative of a function - I see no connection.

The condition is satisfied at three points in the given domain.

 

[Ryan Rigdon]

my bad. let me start from the beginning. i have been up for two days straight and am tired as ****.

the problem.

The Mean Value Theorem applies to the given function on the given interval. Find all possible values of c. Then sketch the graph of the given function on the given interval.

s(?) = 8cos(?) ; [-?,?]

my work thus far

Evaluate s(?) & s(-?)

S(?) = 8cos(?) = -8 s(-?) = 8cos(-?) = -8

Sub values into (s(b) - s(a))/(b-a) & simplify. a= -? b= ?


(-8+8)/(?+?) = 0

Now apply the Mean Value Theorem

sub things back into

(s(b)-s(a))/(b-a) = s'(c)

0 = -8sin(?)

sin(?) = 0

sin^(-1) (0) = c

Values of c = ? (this is where i am stuck)

 

[galactus]

The values that satisfy \(\displaystyle sin(x)=0\) are \(\displaystyle x=C{\pi}\)

Where C is a integer constant \(\displaystyle (-\infty,....-1, 0, 1, ....,\infty)\). You are covering the interval \(\displaystyle [-\pi, \;\ \pi]\), so what values

of C give results in this interval?.

 

[Ryan Rigdon]

with the choices i have been given

A. -?,-?/2,0,?/2,? B. ?/2,-?/2 C. 0 D. No values of C in the given domain satisfy the theorem.


C is my answer

its the only choice that gives me sin(?) = 0.


Does this give the same answer as sin^(-1) (0) = c

 

[BigGlenntheHeavy]

\(\displaystyle arcsin(0) \ = \ c \ implies \ sin(c) \ = \ 0, \ -\pi \ \le \ c \ \le \ \pi\)

\(\displaystyle Hence, \ c \ = \ -\pi, \ 0, \ \pi\)

 

[Ryan Rigdon]

\(\displaystyle arcsin(0) \ = \ c \ implies \ sin(c) \ = \ 0, \ -\pi \ \le \ c \ \le \ \pi\)

\(\displaystyle Hence, \ c \ = \ -\pi, \ 0, \ \pi\)

that is true but I was only given the answer from my choices of O. So i chose C. got it right.