I don't know how to go about this problem
Find the x value at which f is not continuous. Is the discontinuity removable?
f(x)= -6x, x less than or equal to 4
x^2-3x+9 x greater than 4
When x is less than 4, the function of f(x)=?
Is the problem that you do not know what these words
mean? A function, f, is "continuous at x= a" if and only if both f(a) and \(\displaystyle \lim_{x\to a} f(x)\)
exist and \(\displaystyle \lim_{x\to a} f(x)= f(a)\).
(Since the last implies the first two, often only the last, \(\displaystyle \lim_{x\to a} f(x)= f(a)\) is given.)
You should also know that any polynomial, and so f(x)= -6x and f(x)= x^2- 3x+ 9, is continuous for all x. So this f has only one point, x= 4, where it
might not be continuous. \(\displaystyle \lim_{x\to a} f(a)\) exist if and only if the two one-sided limits, \(\displaystyle \lim_{x\to a^-}f(x)\) and \(\displaystyle \lim_{x\to a^+} f(x)\) exist and are equal.
Here, f(x)= -6x for all x< 4 so \(\displaystyle \lim_{x\to 4^-} f(x)= \lim_{x\to 4} -6x\). What is that limit? Also f(x)= x^2- 3x+ 9 for all x> 4 so \(\displaystyle \lim_{x\to 4^+}= \lim_{x\to 4} x^2- 3x+ 9\). What is that limit? Are they equal? If so it that common limit equal to f(4)?