# Thread: help with this calc problem - differentiation, continuity, has variables

1. ## help with this calc problem - differentiation, continuity, has variables

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

i know the basic process: solve for x, set the limits equal, differentiate those 2 equations. but i get stuck with all the variables.
thanks!!

EDIT** yeah youre right i fixed it. but can anyone help me answer it?

2. are you sure that is mx+a if x>0? I think that should say x>1

3. $\displaystyle y = 2x^2 + 3ax +1$ if $\displaystyle x \leq 1$

$\displaystyle y = mx+a$ if $\displaystyle x>1$

for continuity ...

$\displaystyle 2(1)^2 + 3a(1) + 1 = m(1) + a$

for differentiability ...

$\displaystyle 4(1) + 3a = m$

solve the system for a and m.

4. In general this problem solve whit skeeter method, only you know is that if the function is continue the lateral limit are equals