Find the Values of a and b in a piecewise function. Help please

[jbender]

Find the values of a and b that make the function f(x) differentiable.



f(x)={ ax^3 + 1 , for x<2
{ b(x-3)^2 +10, for x is greater than or equal to 2



ax^3 + 1 = b(x-3)^2 +10
3ax^2 = 2b(x-3)
From here I'm unsure of where to go.

 

[Mark44]

Find the values of a and b that make the function f(x) differentiable.



f(x)={ ax^3 + 1 , for x<2
{ b(x-3)^2 +10, for x is greater than or equal to 2



ax^3 + 1 = b(x-3)^2 +10
3ax^2 = 2b(x-3)
From here I'm unsure of where to go.

The two separate definitions are both continuous and differentiable on their separate domains, so you want to find a and b so that these functions are continuous and differentiable at the point where the definitions change. Where is that?

 

[jbender]

The point where the definitions change is at x=2. Correct?

 

[Mark44]

Yes, correct.

 

[HallsofIvy]

That's certainly a good start! You have used the fact that the function must be continuous at x= 2 and the derivatives must "match up". (A derivative is not necessarily continuous but it must satisfy the "intermediate value property" so that is correct. But the point is that must be true at x= 2. Set x= 2 in your equations and you will have two linear equations to solve for a and b.

 

[jbender]

This is my work for setting x=2 in the two given equations.
a(2)^3 +1 = b(2-3)^2 +10
8a+1 = b+10
b=8a-9

What equation do I substitute "b=8a-9" to solve for the value "a" ?

 

[SammyS]

f '(2) must exist. That means that \lim_{x\to2^+}f&#039;(x)=\lim_{x\to2^-}f&#039;(x)=f&#039;(2)\,.

Differentiate f(x) in both regions, x > 2, and x < 2 .