# What is the domain and range of  f(t) = root3(3) sqrt (6t − 2)?

Nov 10, 2017

Domain: $t \ge \frac{1}{3} \mathmr{and} \left[\frac{1}{3} , \infty\right)$
Range : $f \left(t\right) \ge 0 \mathmr{and} \left[0 , \infty\right)$

#### Explanation:

$f \left(t\right) = \sqrt[3]{3} \sqrt{6 t - 2}$ Domain: Under root $\ge 0$ otherwise

$f \left(t\right)$ will be undefined. $\therefore 6 t - 2 \ge 0 \mathmr{and} t \ge \frac{1}{3}$.

Domain: $t \ge \frac{1}{3} \mathmr{and} \left[\frac{1}{3} , \infty\right)$ . Range will not be any negatve

number , so Range : $f \left(t\right) \ge 0 \mathmr{and} \left[0 , \infty\right)$

graph{3^(1/3)*sqrt(6x-2) [-20, 20, -10, 10]}