Question

How do you find the intervals on which the function is continuous given $y = \ln (3x - 1)$?

Answer 1

We use the concept of continuity of logarithm function and write the condition for value of dependent variable in the logarithm function i.e. define the domain of the log function. Use the domain to write the domain for a given function.

Complete step by step answer: We have to find the interval on which the function $y = \ln (3x - 1)$ is continuous.
A function is said to be continuous on a given interval if it is continuous on each point of the interval. Also, we can say a function is continuous in an interval if its graph is a smooth curve i.e. doesn’t break at any point between the intervals.
Now we know that for a logarithm or ln function, the function is only defined for positive real numbers. So, the function is continuous at every point i.e. a positive real number.
So, we can write the function $\log (x)$ or $\ln (x)$ is continuous for every $x > 0$
Here we are given the function $y = \ln (3x - 1)$
So, for the given function to be continuous we have to have $(3x - 1) > 0$
Solve the inequality by shifting constant value to right hand side of the inequality
$\Rightarrow 3x > 1$
Divide both sides of the inequality by 3
$\Rightarrow \dfrac{{3x}}{3} > \dfrac{1}{3}$
Cancel same factors from numerator and denominator on both sides of the inequality
$\Rightarrow x > \dfrac{1}{3}$
So, for the function $y = \ln (3x - 1)$to be continuous, x should be greater than $\dfrac{1}{3}$i.e. the condition on the lower limit and there is no limit or condition on the upper limit.
So, the function$y = \ln (3x - 1)$is continuous for $x \in \left( {\dfrac{1}{3},\infty } \right)$
$\therefore$The interval on which the function $y = \ln (3x - 1)$ is continuous is $\left( {\dfrac{1}{3},\infty } \right)$.

Note:
Many students make mistake of writing the interval for the function as $\left( {1,\infty } \right)$which is wrong, they think logarithm function is continuous for every ‘x’ as positive real number but here if we put in the positive real value less than $\dfrac{1}{3}$we will get the function discontinuous. Keep in mind the independent variable inside the function decides the interval as we write the value of the independent variable greater than 0.