# Continuity Question

1. Dec 26, 2011

### Miike012

A function f is defined as follows:

f(x) = 2cos(x) if x≤c,
= ax^2 + b if x > c .

Where a,b, and c are constants. If b and c are given. find all values of a for which f is continous at the point x = c

Solution:
a = (2cos(c) - b)/c^2 if c ≠ 0 ; if c = 0 there is no solution unless b = 2.

I dont understand how if c = 0 and b = 2 there is a solution at c = 0

For instance... (2cos(c) - 2)/c^2 = (2/c)( Cos(c) - 1 )/c = 2/0*0 as c → 0 which is still a discontinuity.

2. Dec 26, 2011

### gb7nash

In order for f(x) to be continuous at x = 0, f(0) must be defined and:

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$$

Looking at the limit part of this, calculate $\lim_{x \to 0^-} f(x)$. What do you get? Now look at $\lim_{x \to 0^+} f(x)$. What is the only value of b that we can have to get the right limit?

3. Dec 26, 2011

### SammyS

Staff Emeritus
Generalizing on what gb7nash said: f is continuous at at x=c if and only if
$\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$​
Of course since cos(x) is continuous on ℝ, $\displaystyle \lim_{x \to c^-}f(x)=f(c)$ is true for all real c.

$\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)$ leads directly to the equation
$\displaystyle 2\cos(c)=ac^2+b$​
Use that last equation to solve the two cases:
(1) c = 0

(2) c ≠ 0​