Thanks in advance
While it is true that ln(x) > 0 and ln(x²) = 2ln(x), the question did ask for the intercept of ln(x²), not 2ln(x). There is a difference, for example has no y intercept but if you simplify it to y = x it does have one.To find the y intercept, use the fact that ln(1)= 0. f(1)= 2(1)ln(1)= 0. If x> 1, 2x ln(x)> 0 and if x< 1, 2xln(x)< 0 so that is the only y-intercept.
In which case we also need to consider the negative values of x because it is squared to become positive within the natural logarithm function.
This is because ln(0) is undefined. the point (0,0) becomes a point of discontinuity. When graphing, you draw a hollow circle around the point to indicate this.but when i graph it x=0 doesnt exist! why is this??