Question

Find a and b such that the derivative of f exists everywhere, if

f(x)= {x^2+3, (x < or =1) and ax^3+bx+4, x>1

Answer 1

this is much easier than it looks
take the derivative of both
replace \(x\) by \(1\) and set them equal

Answer 2

so 2x
and
a3x^2+b

2(1)=1
a3(1)^2+b=1

?

Answer 3

lost.

Answer 4

still trying to figure out if I am on the right track. I am lost right now. No idea what to do after I find their derivatives never had an a and b in the equation like this.

Answer 5

@Fakshon
as far as derivatives are concerned...you got that part right..
2x
and
3a*x^2+b
now you have to set:
3a*x^2+b=2x
and put x=1

Answer 6

so you got one equation with a and b?

Answer 7

still lost?Just nod...we can go to the begining

Answer 8

you function is defined in 2 different ways depending on the interval. The limit point of the intervals is x=1. Independently, the derivative exist on each interval separatly, but for them to exist in all R it is needed that they are equal in this limiting point x=1. So like @satellite73 said, find first each derivative. Later set x=1 and put them equal. What is left is to choose a and b that make the equality happen. That's it

Answer 9

@myko
we get one equation in a and b by differentiating both parts and setting them equal at x=1...the other equation is obtained from the fact that function is continuous at x=1...since continuous functions are differentiable and vice versa...
so we can equalize x^2+3 and ax^3+bx+4 at x=1..so that the function is continuous..
now two equations..two unknowns..
can be worked out

Answer 10

right