let f(x)= bx^2+6x, if x< or equal to 2

Question

ax^3, if x>2
find the values of a and b such that f(x)is differentiable at x=2

I need help with this question I know you have to set both functions equal to each other. but I don't know how to solve for a and b

Answer 1

f(x) =
bx^2+6x if x <= 2
ax^3 if x>2

So, we have
lim(x->2-) f(x) = 4b+12
lim(x->2+) f(x) = 8a
That means we need
4b+12 = 8a

Also, we need the derivatives to be equal:
2bx+6 -> 4b+6
3ax^2 -> 12a
And they need to be equal: 4b+6 = 12a

rearranging things a bit, we have

8a-4b = 12
12a-4b = 6

a = -3/2
b = -6

f(x) =
-6x^2+6x if x <= 2
-3/2 x^3 if x > 2

Check the graphs. There should be a smooth transition at x=2:

http://www.wolframalpha.com/input/?i=plot+y+%3D+-6x^2%2B6x%2C+y+%3D+-3%2F2+x^3+for+x+%3D+-1..3