how to find the value of k

Question

f(x)=\begin{cases} k(x^2-2x),x\le 0 \\ 4x+1,x>0 \end{cases} continuous at x=0

I am extremely weak in this topic so could any one show me how to solve this question?

Answer 1

You need to make sure that the different pieces of the function meet at the same value at $x = 0$.

For them to meet at the same value at $x = 0$, we need $k(0^{2} - 2(0)) = 4(0 + 1)$.

This simplifies to $k*0 = 4$. But for any $k$, this gives $0 = 4$. But $0$ never equals $4$... So since we arrived at a false statement, and this false statement came up no matter which $k$ we chose, we can conclude that there are no values of $k$ that make the function continuous at $x = 0$.

Answer 2

$f(x)$ is said to be continuous at $x=a$ if

$$\lim_{x\to a^-}f(x)=f(a)=\lim_{x\to a^+}f(x)$$