# Continuity question

#### db5vry

The function f is defined by

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

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

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?

Thanks for your help #### Plato

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

• db5vry

#### SpringFan25

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:

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

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

$$\displaystyle 3a +2b = 2+1$$

$$\displaystyle 3a +2b = 3$$

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

$$\displaystyle a + b = 2$$

• db5vry

#### db5vry

From the given we get this system:
$$\displaystyle a+b=2~\&~3a+2b=3$$.
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 #### SpringFan25

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.

• HallsofIvy and db5vry

#### db5vry

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 