prelim Domain & Range of inverse trig function (1 Viewer)

[fx82au]

anyone know how to find the range for 3a?

I can find the domain using inequalities, but not the range...

the answer in the book is [0, pi/2]

thanks in advance

 

[cossine]

-1 <= x^2 <= 1

=>

-1 <= x <= 1

You need to take an inspection based approach for calculating range. The smallest value of x^2 is 0 and the largest is 1.

 

[fx82au]

and the largest is 1

how did you get that? from the graph?

 

[cossine]

how did you get that? from the graph?

By inspection

-1 <= x <= 1

=> The smallest value of x^2 is 0 and the largest is 1

You need to examine the domain.

 

[fx82au]

By inspection

-1 <= x <= 1

=> The smallest value of x^2 is 0 and the largest is 1

You need to examine the domain.

sorry, im not sure I understand if youre talking about the domain or range. how do i get the range using

an inspection based approach for calculating range

thanks

 

[cossine]

I will try to explain in words.

So if (x, y) is a point in function f. Then (y, x) is a point on f^-1. The unit circle definition of sin is sin theta = y.

The domain is given by -1 <= x <= 1. Using this domain we can determine the range of x^2 which in turn help us determine the range of sin^-1(x^2)
Since the smallest value x^2 is 0 and the largest value is 1. Try substituting x=0, 1 and you will see 0<= x^2 <=1.

Since sin^-1(x) is a monotonically increasing function. By substituting the values x^2 =0 and x^2 = 1 you will get the range.

sin^-1(0) = 0 # verification sin(0) = 0
sin^-1(1) = pi/2 # verification sin(pi/2) =1

Therefore the range is [0, pi/2]

Note: I presume you already know arcsin is same as sin^-1.