Question #12cba

Jun 5, 2016

The function is defined and continuous on $x > \frac{3}{2}$.

Explanation:

The function $y = \ln \left(x\right)$ is only defined when $x > 0$. The function is continuous on its entirety when it is defined.

So, for the function $y = \ln \left(4 x - 6\right)$, we know that $4 x - 6 > 0$. This can be solved to show that $x > \frac{3}{2}$.

Therefore the function $y = \ln \left(4 x - 6\right)$ is defined on the interval $x \in \left(\frac{3}{2} , \infty\right)$ and is continuous on that entire interval.

The graph of $y = \ln \left(x\right)$:
graph{ln(x) [-9.1, 36.51, -12.19, 10.63]}

The graph of $y = \ln \left(4 x - 6\right)$:
graph{ln(4x-6) [-9.1, 36.51, -12.19, 10.63]}