# Finding a value to make piecewise continuous

1. Feb 2, 2009

### BrianHare

1. The problem statement, all variables and given/known data

Find c such that it makes f(x) continuous.

2. Relevant equations

$f(x)=\begin{cases} 2x+c&x < -5\\ 3x^2&-5 \leq x < 0\\ cx^2&0 \leq x\\ \end{cases}$

3. The attempt at a solution

I know that
$$\lim_{x\to 5^-}3x^2$$ = 2x+c
and
$$\lim_{x\to 0^+}3x^2$$ = cx^2

Which makes the 2 points where it disconnects at (-5, 75) and (0,0)
Given that I make 2x+c = 75, where x=-5. I get C= 85, and since cx^2=0 is any real, does that mean the answer is c=85?

Last edited: Feb 2, 2009
2. Feb 2, 2009

### tiny-tim

Welcome to PF!

Hi Brian! Welcome to PF!

Yes, c = 85.

(though you have a strange way of using lim …

you might as well say lim 3x2 = 3*(-5)2, and so on. )

3. Feb 2, 2009

### BrianHare

Re: Welcome to PF!

My teacher has a definition where

$$\lim_{x\to a}f(x)$$ = f(c)

It was my understanding that f(x) = 3x^2, a = -5, and f(c) = 2x+c. So once I knew the answer to the limit, i knew that f(c) = 75, thus 2x+c must also be 75. Maybe I am misunderstanding the definition. Can anyone clarify?

4. Feb 2, 2009

### tiny-tim

that doesn't make any sense …

what is f(c) supposed to mean?

(f(c) = 2c + c or 3c2 or cc2)

does he mean $$\lim_{x\to a}f(x)$$ = f(a)?​