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

May 31, 2018

$2 \le y < \infty$

#### Explanation:

Given ${\log}_{0.5} \left(3 x - {x}^{2} - 2\right)$

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:

$- {x}^{2} + 3 x - 2 = 0$

${x}^{2} - 3 x + 2 = 0$

$\left(x - 1\right) \left(x - 2\right) = 0$

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

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

$y = {\log}_{0.5} \left(3 x - {x}^{2} - 2\right)$

Convert the base to the natural logarithm:

$y = \ln \frac{- {x}^{2} + 3 x - 2}{\ln} \left(0.5\right)$

To find the minimum, compute the first derivative:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x + 3}{\ln \left(0.5\right) \left(- {x}^{2} + 3 x - 2\right)}$

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

$0 = \frac{- 2 x + 3}{\ln \left(0.5\right) \left(- {x}^{2} + 3 x - 2\right)}$

$0 = - 2 x + 3$

$2 x = 3$

$x = \frac{3}{2}$

The minimum occurs at $x = \frac{3}{2}$

$y = \ln \frac{- {\left(\frac{3}{2}\right)}^{2} + 3 \left(\frac{3}{2}\right) - 2}{\ln} \left(0.5\right)$

$y = \ln \frac{\frac{1}{4}}{\ln} \left(0.5\right)$

$y = 2$

The minimum is 2.

Because $\ln \left(0.5\right)$ is a negative number, the function approaches $+ \infty$ as x approaches 1 or 2, therefore, the range is:

$2 \le y < \infty$