1. Differentiable equations

The problem: Find the value of b that makes the function differentiable.

f(x) =

ax^2 + 10, x> 2

x^2 -6x +b, x < 2

It is a peicewise function just incase that doesn't look right to someone.

I have attempted setting both functions equalto one another but all I ended up with was b-4a = 2 and I am stuck after that.

2. the function needs to be continuous ...

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)$

$4 - 12 + b = 4a + 10$

and so does the derivative of f ...

$\lim_{x \to 2^-} f'(x) = \lim_{x \to 2^+} f'(x)$

$4 - 6 = 4a$

3. so then a = -1/2?

but that makes my b = -1 and for what Im doing it is not possible to have negative answers. Am I just wrong in my calculations?

Yes I am wrong, I multiplied by -8 instead off subtracting.

4. Originally Posted by CalcGeek31
so then a = -1/2?

but that makes my b = -1 and for what Im doing it is not possible to have negative answers. Am I just wrong in my calculations?
no ... b = 16

5. Yes I got that too thank you for your help.