determine the values of constant real numbers a and b, so that this function is differentiable at x= -5 f(x) = (ax^2) - 4x -76 x 5

Question

determine the values of constant real numbers a and b, so that this function is differentiable at x= -5

f(x) = (ax^2) - 4x -76             x<=(-5)
      = bx - 1                           x > 5

anonymous · Answer

$$ax^2 - 4x - 76,     x \le -5$$
$$bx -1, x > -5$$

anonymous · Answer

answers are  
a = -3
b = 26

anonymous · Answer

what do i do to get the answers though

zepdrix · Answer

Hmm are you sure it's -76? Or is it suppose to be 75?
I get a=-3 if its a 75.. hmm
Sec ill check my work, then I can hopefully explain how i got there.

anonymous · Answer

yes it is -76

zepdrix · Answer

|dw:1349558391356:dw|
Here is a little example.
In order for this piece-wise function be continuous, the limit from the left, needs to EQUAL, the limit from the right (approaching -5).

That's why I drew those little arrows, to show that we're approaching -5 from each side.

Also, their tangent line's need to be approaching the same value at that point -5.
Meaning, their derivatives must be equal when approaching from the left and right.

So we have a little bit of work to do.
And it will give us a system of 2 equations, and we'll be able to solve for a and b from there.