How do you find the domain and range of f(x) =sqrt( x - 1) - 1/sqrt( 2-x)?

1 Answer
Monzur R.
May 30, 2017

1 <= x< 2
f(x) <= f(1.318)

Explanation:

We have f:x|->sqrt(x-1)-1/sqrt(2-x)

Whenever we have a function with a square root, we know that the square root has to be bigger than or equal to 0.

sqrt(x-1) >= 0

x-1 >= 0

x >= 1

For the second square root, we need to be more careful, because we have 1"/"sqrt(2-x). This means that sqrt(2-x) > 0.

2-x > 0

2 > x

So our domain is 1 <= x < 2

To find the range, we need to find the stationary point and find if the function is increasing or decreasing.

f'(x)=1/2(1/sqrt(x-1)-1/(2-x)^(3/2))

Since 1"/"(2-x)^(3/2) > 1"/"sqrt(x-1), f'(x) < 0 so the function is decreasing. This means the function will have a maximum point, and all other y-values will be below that point.

Let f'(x)=0

1/2(1/sqrt(x-1)-1/(2-x)^(3/2))=0

Solving for x, we get x=1.318.

Therefore, f(x) <= f(1.318)

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