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

1 Answer
May 31, 2018

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