# Help with calculus problem- differentiability, continuity, with variables

1. Nov 2, 2008

### meredith

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

y=
y= {1+3ax+2x^2} if x is < or = 1
{mx+a} if x>0
what values for m and a make x continuous and differentiable at 1?

2. Relevant equations
n/a

3. The attempt at a solution
i solved for when x=1.
i got 3+3a.
this is also the right hand limit. i kow you have to set the two limits equal for it to be continuous. so i got 3a+3=m+a. m=2a+3.
but honestly i dont know where to go from there. i know that i have to take the derivative of those two equations too. but i dont even kow what to say too many variables!
mer

2. Nov 2, 2008

### phyzmatix

For $$f(x)$$ to be continuous at

$$x=a$$

we must have

$$\lim_{x\rightarrow a}f(x)=f(x)$$

So, using your system of inequalities, you need to have

$$f(a)=\lim_{x\rightarrow a^-}f(x) \mbox{(i.e. where f(x) < a)}=\lim_{x\rightarrow a^+}f(x)\mbox{(i.e. where f(x) > a)}$$

(NOTE: From left to right, the weird semi-colon and the upside down question mark are supposed to denote "smaller than" and "bigger than" symbols respectively. Don't know why Latex is playing with me.)

Does that help?

Last edited: Nov 2, 2008
3. Nov 2, 2008

### HallsofIvy

You haven't yet used the requirement that the function be differentiable at x= 1. While a derivative does not have to be continuous itself, it must satisfy the "intermediate value property" which, among other things, means that, as long as the derivative exists, it must be equal to the right and left hand limits.
The derivative of 1+ 3ax+ 2x2 is 3a+ 4x and, at x= 1, that is 3a+ 4. The derivative of m so, in order that this function be differentiable at x=1, you must have 3a+ 4= m. That, together with m= 2a+ 3 gives you 2 equations to solve for m and a.