Help with calculus problem- differentiability, continuity, with variables

[meredith]

Homework Statement

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?

Homework Equations

n/a

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 don't know where to go from there. i know that i have to take the derivative of those two equations too. but i don't even kow what to say too many variables!
thanks in advance!
mer

 

[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) &lt; a)}=\lim_{x\rightarrow a^+}f(x)\mbox{(i.e. where f(x) &gt; 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?

 

[HallsofIvy]

Homework Statement

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?

Homework Equations

n/a

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 don't know where to go from there. i know that i have to take the derivative of those two equations too. but i don't even kow what to say too many variables!
thanks in advance!
mer

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.