## Precalculus (6th Edition) Blitzer

a) $\left( -\infty ,0 \right)\cup\left( \text{1}\text{.5,3} \right)$ b) $\left( 0,1.5 \right)\cup\left( 3,\infty \right)$ c) nowhere
(a) The function is said to be increasing in an interval when the value of y increases as the value of x increases. So the value of y is increasing with an increase in the value of x on the intervals $\left( -\infty ,0 \right)\cup\left( \text{1}\text{.5,3} \right)$. (b) Since the function is said to be decreasing in an interval when the value of y decreases as the value of x increases. So the value of y is decreasing with an increase in the value of x on the intervals $\left( 0,1.5 \right)\cup\left( 3,\infty \right)$ (c) The function is said to be constant in an interval when the value of y remains the same as the value of x increases. The value of y is nowhere constant with an increase in the value of x. Hence, the given function is constant nowhere.