Implied domain and range

[Zebra]

We are told that the domain of Cos(x) is 0 to pie

Cos(sin-1(x))

the implied domain?

Going through questions I got wrong in Exercise 3C is driving me mad!

[xZero]

since the range of sin^-1(x) is -pi/2 to pi/2, for the function Cos(sin^-1(x)) to exist, we must restrict the domain of sin^-1(x) to x=0 to x=1. Hence the implied domain is 0≤x≤1

[Vincezor]

So, we need find the implied domain of

as you said,

You may remember from composite functions in methods that

=

However we must restrict this range so that it becomes a subset of the domain of cos(x), so we get

0 ≤ ≤

EDIT: Solve for x

sin(0) ≤ x ≤ sin()

which is

0 ≤ x ≤ 1



Is this working out right? :S

[luffy]

So, we need find the implied domain of

as you said,

You may remember from composite functions in methods that

=

However we must restrict this range so that it becomes a subset of the domain of cos(x), so we get

0 ≤ ≤

Multiply everything by so middle becomes x

sin(0) ≤ x ≤ sin()

which is

0 ≤ x ≤ 1



Is this working out right? :S

Yes - ur working is correct. Except, your not multiplying everything by sin....

[Vincezor]

Yes - ur working is correct. Except, you're () not multiplying everything by sin....

Oh my bad, was I meant to say I was solving for x? Because I think that's what I was meant to do...

[luffy]

Yes - ur working is correct. Except, you're () not multiplying everything by sin....

Oh my bad, was I meant to say I was solving for x? Because I think that's what I was meant to do...

LOL! - Haha... I actually laughed at how you edited my quote. Good one!

P.S. I would love to know you in real life