Find Constant $A $ and $B$

Question

$$f(x)=\begin{cases} \frac{2x^2-3x+1}{2x-1} &, x<\frac{1}{2}\\ Ax+B &, x=\frac{1}{2}\\ 2B &, x>\frac{1}{2}\end{cases}$$ Determine constants $A$ and $B$ such that $f(x)$ is continuous for all values of $x$. Show your work using the conditions of continuity.

So the 1st function can be factored into $(x-1), x \lt 1/2$

And $f(1/2) = Ax+B$

But I can't quite find the value of $A$ and $B$. I found many examples related to this but nothing could help me out. If you guys could help me out in any way that would be great.

Answer 1

SOLUTION:

Analyzing left continuity at $\frac{1}{2}$, using L'Hopital's Rule:

$$Ax+B=\frac{4*\frac{1}{2}-3}{2}$$ $$\frac{A}{2}+B=\frac{-1}{2}$$

Lets check the right continuity:

$$\frac{A}{2}=B$$

From the two equations, we obtain $B=\frac{-1}{4}$ and $A=\frac{-1}{2}$

Answer 2

Just compute $$\lim_{x\to\frac{1}{2}^-}f(x)=f(\frac{1}{2})=\lim_{x\to\frac{1}{2}^+}f(x)$$

Answer 3

${\Large HINT:}$

If a function is continuous at some point, say $x=1/2$ for this question, then the $$ \lim_{x \to (1/2)^{+} } f(x) = \lim_{x \to (1/2)^{+1}} f(x) = f(1/2) $$

${\Large UPDATE:}$.

It doesn't make sense for $f(1/2) = Ax +B$.

Answer 4

The limits of the function pieces at $x=\frac12$ (in their respective domains) must be equal and equal $f(\frac12)$.

$$f\left(\tfrac12^-\right)=f\left(\tfrac12\right)=f\left(\tfrac12^+\right),$$

$$-\frac12=\frac12A+B=2B.$$