What is the range of f(x) = ln(sin^(-1)(x^2+x+3/4))f(x)=ln(sin−1(x2+x+34)) ?

1 Answer
George C.
Jun 15, 2017

[ln(pi/6), ln(pi/2)] ~~ [-0.6470, 0.4516][ln(π6),ln(π2)]≈[−0.6470,0.4516]

Explanation:

Given:

f(x) = ln(sin^(-1)(x^2+x+3/4))f(x)=ln(sin−1(x2+x+34))

First note that:

x^2+x+3/4 = x^2+x+1/4+1/2 = (x+1/2)^2+1/2x2+x+34=x2+x+14+12=(x+12)2+12

which can take any value in the range [1/2, oo)[12,∞)

The domain of sin^(-1)sin−1 as a real valued function of real arguments is [-1, 1][−1,1].

So the possible valid arguments to it in f(x)f(x) are all in:

[1/2, oo) nn [-1, 1] = [1/2, 1][12,∞)∩[−1,1]=[12,1]

sin^(-1)(1/2) = pi/6sin−1(12)=π6

sin^(-1)(1) = pi/2sin−1(1)=π2

and sinsin is monotonically increasing between these two endpoints.

So the range of sin^(-1)(x^2+x+3/4)sin−1(x2+x+34) is [pi/6, pi/2][π6,π2]

Hence the range of f(x)f(x) is:

[ln(pi/6), ln(pi/2)][ln(π6),ln(π2)]

Related questions
See all questions in Domain and Range of a Function
Impact of this question
1245 views around the world
You can reuse this answer
Creative Commons License