$\displaystyle f(x)=xln(x^2)$
Thanks in advance
Since $\displaystyle ln(x^2)= 2ln(x)$ that is the same as $\displaystyle f(x)= 2xln(x)$. There is no x intercept because ln(0) is not defined. 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.
so when $\displaystyle f(x)=0$
it's $\displaystyle 0=xln(x^2)$
Therefore $\displaystyle 0=x$ and $\displaystyle 0= \ln (x^2) $
$\displaystyle 0= \ln(x^2) $
$\displaystyle \Rightarrow e^0 = e^{\ln (x^2)} $
$\displaystyle \Rightarrow 1=x^2 $
$\displaystyle \Rightarrow x=-1, 1 $
but when i graph it x=0 doesnt exist! why is this??
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 $\displaystyle y = \frac{x^2}{x} $ 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??