Question

Range of `log_0.5(3x-x^2-2)` ?

Answer 1

`2 <= y < oo`

Explanation:

Given `log_0.5(3x-x^2-2)`

To understand the range, we need to find the domain.

The restriction on the domain is that argument of a logarithm must be greater than 0; this compels us to find the zeros of the quadratic:

`-x^2+ 3x-2 = 0`

`x^2- 3x+2 = 0`

`(x -1)(x-2) = 0`

This means that the domain is `1 < x < 2`

For the range, we set the given expression equal to y:

`y = log_0.5(3x-x^2-2)`

Convert the base to the natural logarithm:

`y = ln(-x^2+3x-2)/ln(0.5)`

To find the minimum, compute the first derivative:

`dy/dx = (-2x+3)/(ln(0.5)(-x^2+3x-2))`

Set the first derivative equal to 0 and solve for x:

`0 = (-2x+3)/(ln(0.5)(-x^2+3x-2))`

`0 = -2x+3`

`2x=3`

`x = 3/2`

The minimum occurs at `x = 3/2`

`y = ln(-(3/2)^2+3(3/2)-2)/ln(0.5)`

`y = ln(1/4)/ln(0.5)`

`y = 2`

The minimum is 2.

Because `ln(0.5)` is a negative number, the function approaches `+oo` as x approaches 1 or 2, therefore, the range is:

`2 <= y < oo`