1. ## Another Differentiable Equation

Okay this one is different. when I take the derivative I assume I can just plug in but I am not sure if it will work.
f(x)=
$ax^2 + 1/3$, x > 1

$bx-10/3$, x <1

I took the limit, and got $a +1/3 = b - 10/3$, then I took the derivative of the equation and took the same limit and got 2a = b. Do I just plug 2a in for b?

2. I am getting 11 but it would be very helpful to me if someone would explain it.

3. Originally Posted by CalcGeek31
Okay this one is different. when I take the derivative I assume I can just plug in but I am not sure if it will work.
f(x)=
$ax^2 + 1/3$, x > 1

$bx-10/3$, x <1

I took the limit, and got $a +1/3 = b - 10/3$, then I took the derivative of the equation and took the same limit and got 2a = b. Do I just plug 2a in for b?
What's the original question? Are you being asked to find the values of a and b so that curve is differentiable at x = 1? If so, then what you propose is correct.

By the way, is $ax^2 + 1/3$ meant to be $ax^2 + \frac{1}{3}$ or $\frac{ax^2 + 1}{3}$ ....?

Originally Posted by CalcGeek31
I am getting 11 but it would be very helpful to me if someone would explain it.
11 for what?