# this is math spec , i need help and then i will try , review

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gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

At enotes, the educators help students work towards their problems, rather than solving their complete assignments for them. So, I will solve some of your problems and you can use them as examples and solve the rest yourself.

a) i) f(3) = `sin(2tan^(-1)(3/3)) = sin (2tan^(-1)(1)) = sin (2 pi/4) = sin(pi/2) = 1`

(ii) `f(-sqrt(3)) = sin(2tan^(-1)(-sqrt3 /3)) = sin(2tan^(-1) (-1/sqrt(3))) `

`= sin(2 * -pi/6) = sin(-pi/3) = -sqrt(3)/2`

b) since the domain of sin function is restricted to [-pi/2, pi/2], the domain of f(x) will also be `[-pi/2, pi/2]`

c) Range of f(x): In the given domain, sin function has a range of [-1,1].

thus the implied range of f(x) is also [-1,1].

f(2) can be calculated in the same way as f(3) was calculated.

`f^(-1) (x)` can be calculated by solving for x and then replacing x by y.

if we solve the function, f(x) = y (say),

we get, `x = 3 tan ((sin^(-1)y)/2)`

and then we replace x by y and vice versa, we get

`f^(-1) (x) = 3 tan((sin^(-1)x)/2)`

substituting x = -3/5 in the inverse function, we get

`f^(-1) (-3/5) = -1`

Try doing rest of the parts on your own, using the above as examples.

hope this helps.