Continuity question

Oct 2008
157
3
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
Aug 2006
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From the given we get this system:
\(\displaystyle a+b=2~\&~3a+2b=3\).
 
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May 2010
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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
\(\displaystyle \lim_{x \to 1^{-}} ax^{3}+bx^{2} = \lim_{x \to 1^{+}} x^{2} + x \)

\(\displaystyle a + b = 2 \)
 
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Oct 2008
157
3
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 :)
 
May 2010
1,034
272
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.
 
Oct 2008
157
3
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 :)