Help finding the values of a,b and c.

Question

f(x) = ax^2 + bx         when x<-1
       = 3                      when x=-1
       = x^3 - cx            when x>-1.

What values of a,b and c will ensure that f is continuous at x=-1. What values will ensure that f will be differentiable at x =-1.

I got the value of C to be four as it can be found simply by making the equation equal to 3. However that cannot be done for the first function as it has two pro numerals and only one known point, so I'm stuck. Help would be appreciated.

Answer 1

The key clue here is that the resulting curve must be differentiable, i.e., the slope of the lines on either side of the connection point at (-1, 3) must have the same slope. Having resolved the value of c as +4, in the third piece, we can differentiate that piece. We should then get:

 

       f'(x) = 3x² - 4         when x > -1

 

Therefore, at x = -1

 

      f'(-1) = 3 - 4 = -1

 

The first piece of the function has two parameters to be resolved, a and b. If we evaluate the function at the point (-1, 3), we get:

 

      f(-1) = a(-1)² + b(-1) = a - b = 3

so

      a = b + 3

 

If we differentiate the function's first branch, we get:

 

      f'(x) = 2ax + b

 

and we need to set that to -1 at x = -1 to have the curve continuous and differentiable at the join.

 

So we now have two equations for the first piece of the function:

 

      -2a + b = -1

      a = b + 3

 

this will give us b = -5 and so a = -2

 

Placing these values into the equations, we get the full function being:

 

      f(x) = -2x² - 5x          when x < -1

            = 3                      when x = -1

            = x³ - 4               when x > -1

 

This should be differentiable with a slope of -1 at x = -1, and continuous through the point (-1, 3).