# Removable continuity

1. Dec 17, 2009

### jwxie

This is from my old exam

f(x) = for x <1 (x-1)^2
for 1 <= x <= 4 ax+b find a and b so that fx is continuous for all x
for x <4 sqrt (2x+1)

so i guess i start evaluating some limit.

since the ax+b is define everywhere b/w x = 1 and x = 4, i guess i would use the third function f(x) = sqrt(2x+1) for limit goes to 4

f(x) = limit (x goes to 4) sqrt(s2+1) = 3
then i think i would let a = 3 ????
3x+b = ???? what is y then????

2. Dec 17, 2009

### rochfor1

Where might this function fail to be continuous?

3. Dec 17, 2009

### jwxie

x = 4 and x = 1

4. Dec 17, 2009

### rochfor1

Yes, so what does it mean for this function to be continuous at x = 1 and x = 4?

5. Dec 17, 2009

### jwxie

oh so the ax +b = y
since a = 3, for x = 4, y use the third equation which gives 9
3x+b = 9
am i correct

6. Dec 17, 2009

### rochfor1

Why does a = 3? The definition of continuity is: f is continuous at c if $$\lim_{ x \to c } f(x) = f(c)$$.

7. Dec 17, 2009

### jegues

Shouldn't it be :

for x > 4 : sqrt(2x+1)

8. Dec 18, 2009

### HallsofIvy

Staff Emeritus
Yes, as jeques says, you must mean that $f(x)= \sqrt{2x+1}$ for x> 4.

What is the limit of f as x goes to 1 from below?
What is the limit of f as x goes to 1 from above?

For f to be continuous there, those two limits must be the same.

What is the limit of f as x goes to 4 from below?
What is the limit of f as x goes to 4 from above?

For f to be continous there, those two limits must be the same.

You cannot determine a or b from either of those alone. Each gives an equation in both a and b.