$\begingroup$

I am doing basic calculus, can someone tell me how to find the continuity here at $x=1$

$f(x)=\begin{cases}5x-4 & 0<x \leq 1\\4x^2 -3x & 1 < x < 2.\end{cases}$

$\endgroup$ 5

3 Answers

$\begingroup$

Suppose $a,b\in\Bbb R$ with $a<b$, let $I=(a,b)$, $c\in I$, $f:I\to\Bbb R$. Then we say that $f$ is continuous at $c$ iff $$f(c)=\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x),$$ that is, iff the left and right limits exist, and are both equal to $f(c)$.

In this case, coming toward $c=1$ from the left, what does $f(x)$ look like--that is, how is $f(x)$ defined for $x<1$? What about from the right?

$\endgroup$ 2 $\begingroup$

It's clear that both functions are continuous separately over $\Bbb R$: the problem is continuity at $x=1$. So just make sure that the limits from the right and left of $1$ of$ f(x) = f(1)$.

$\endgroup$ 5 $\begingroup$

The function is continuos because it is both right continuous and left continuos at f(1) where x=1 The limit 5x-4 =1 As x tends to 1 The limit 4x²-3x =1 As x tends to 1 So the function is both left continuous and right continuous hence generally continuous at x=1

$\endgroup$ 0

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy