# Math Help - Continuity question

1. ## Continuity question

The function f is defined by

f(x) = $ax^3 + bx^2$ for $x < 1$
f(x) = $x^2 + x$ for $x \geq 1$

Both f and its derivative are continuous for all values of x. Find the values of the constants a and b.

I find the derivative for both parts as

f'(x) = $3ax^2 + 2bx$ for $x < 1$
f'(x) = $2x + 1$ for $x \geq 1$

I know how to find a and b if they are continuous for one particular value for x - the function and the derivative are solved by means of simultaneous equations.

I tried to do this method for this question, but found that I achieved varying answers - I tried x=1, x=2 and this resulted in different values for a and b, so I'm guessing that this question requires something a bit different? How would you attempt to solve this?

2. From the given we get this system:
$a+b=2~\&~3a+2b=3$.

3. in case some explanation is needed; A function is continuous if its limit from above is the same as its limit from below.

First, apply this condition to the derivative:

$\lim_{x \to 1^{-}} f'(x) = \lim_{x \to 1^{+}} f'(x)$

$\lim_{x \to 1^{-}} 3ax^{2} +2bx = \lim_{x \to 1^{+}} 2x+1$

$3a +2b = 2+1$

$3a +2b = 3$

now, do the same on the main function
$\lim_{x \to 1^{-}} ax^{3}+bx^{2} = \lim_{x \to 1^{+}} x^{2} + x$

$a + b = 2$

4. Originally Posted by Plato
From the given we get this system:
$a+b=2~\&~3a+2b=3$.
And I solved this to get a = -1, b = 3.

But the question said that the function is continuous for all values of x - so shouldn't a and b be the same whatever value of x is used?
I've obviously misunderstood the idea of continuity, but I would like to know how my above statement can be proved wrong if it is

5. Originally Posted by db5vry
And I solved this to get a = -1, b = 3.

But the question said that the function is continuous for all values of x - so shouldn't a and b be the same whatever value of x is used?
I've obviously misunderstood the idea of continuity, but I would like to know how my above statement can be proved wrong if it is

If you look at x values other than 1, the function is continuous for any values of a and b. The only possible point of discontinuity is at x=1 (where the functional form changes) so we only have to inspect it there.

6. Originally Posted by SpringFan25
If you look at x values other than 1, the function is continuous for any values of a and b. The only possible point of discontinuity is at x=1 (where the functional form changes) so we only have to inspect it there.
It makes a lot more sense now seeing that! Thanks for your and everyone's time and help