# Find a and b so that  f(x) = { (x^2, xle1),(ax+b,x gt 1) :}  is continuous?

Mar 27, 2017

$a = 1 - b$

#### Explanation:

We have:

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} & x \le 1 \\ a x + b & x > 1\end{matrix}\right.$

Both ${x}^{2}$ and $a x + b$ are polynomials, and so are continuous in there own right, so the only possibility of a discontinuity is at the junction between the two polynomials at $x = 1$

In order for continuity at $x = 1$ then we require (by the definition of continuity):

${\lim}_{x \rightarrow 1} f \left(x\right) = f \left(1\right)$

Consider the LH limit;

${\lim}_{x \rightarrow {1}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {1}^{-}} {x}^{2} = 1$

And the RH limit:

${\lim}_{x \rightarrow {1}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {1}^{+}} a x + b = a + b$

So the requirement for continuity is:

${\lim}_{x \rightarrow {1}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {1}^{+}} f \left(x\right)$

$\therefore a + b = 1 \implies a = 1 - b$