Inverse functions

The function g : x --> 4-3sinx is defined for the domain 1/2π≤ x ≤A.
1----State the largest value of A for which a has an inverse.
2----For this value of A, find the value of g^-1 (3).

I did 1; A = 3pi/2

But 2, g^-1(x) = sin^-1 ((4-x)/3)
g^-1(3) = sin^-1 ((4-3)/3) = sin^-1 (1/3) = 0.3398 (in radians)
But the answer is 2.80
What am i doing wrong?
Thanks

Reply 1

Zacken

22

Original post by Jmun

The function g : x --> 4-3sinx is defined for the domain 1/2π≤ x ≤A.
1----State the largest value of A for which a has an inverse.
2----For this value of A, find the value of g^-1 (3).

I did 1; A = 3pi/2

But 2, g^-1(x) = sin^-1 ((4-x)/3)
g^-1(3) = sin^-1 ((4-3)/3) = sin^-1 (1/3) = 0.3398 (in radians)
But the answer is 2.80
What am i doing wrong?
Thanks

For the first part, you know

\sin x

repeats itself every

2\pi

radians, so for it to be injective it needs to have a domain that spans 2pi radians or less. (you might have a wrong inequality sign in your OP)

IGNORE. Your answer is correct.

(edited 8 years ago)

Reply 2

Jmun

OP

3

My answer for 1 was correct.

Reply 3

Zacken

22

Original post by Jmun

My answer for 1 was correct.

As far as I can see, your method for 2 is correct except it spits out a value that isn't in g's domain. So, perhaps find an angle equivalent to that. Like

\pi - \arcsin(\frac{1}{3})

Inverse functions

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