Expert: Ahmed Salami - 10/6/2009

Question
QUESTION: Hi, I've been viewing some of the answers to problems that you've been receiving, and thanks to you, I understand them ten times better than I do in class. I hope you will be able to help me with this one.

Let f be the function defined as follows.

f(x)= |x-1|+2, for x<1
     ax^2+bx, for x must be > than or = 1, where a and b are     constants.

(a.) If a=2 and b=3, is f continuous for all x? Justify your answer.

(b.) Describe all values of a and b for which f is a continuous function.

(c.) For what values of a and b is f both continuous and differentiable?

I understand part a, but I'm clueless as to how I should go about b & c. I need this for tomorrow, but frankly, I don't care when you get back to me. As long as I can understand it whenever I do receive a reply, that's all that matters. :)

ANSWER: Hi Devi,
Since you're one for full understanding, lets work together here.
What was your answer to part (a) and why?
Do you think f(x) is continuous at x = 1? Can it be for some values of a and b?

Regards

---------- FOLLOW-UP ----------

QUESTION: My answer for part a was that f was not continuous for all x, because when x=1, the limit does not equal the y value (the limit I believe is 2, whereas f(1)= 5 with the given a & b constants).

And there are a&b values which would allow the function to be continuous.. I'm just not sure how to go about finding those values.. ;D

Thanks so much for working with me. :)

Answer
Hi Devi,
In part (a), f is not continuous for all x because at x = 1 the limit does not equal to the y-value like you have said. But actually, the limit doesnt even exist at all at x = 1 because f doesnt have equal left and right limits. Limit from right is 5 while that from the left is 2.
f would be a continuous function if it is continuous for all x. Clearly it is continuous everywhere else except at x = 1 where continuity is conditional. Now, in part (b) we need to find the values of a and b that makes those limits in part (a) equal. The limit from the left is always 2 since it doesnt depend on the values of a and b. All we need do is find a and b such that the limit as x approaches 1 is equal to 2, i.e
lim (ax² + bx) = 2
x->1
we have,
a + b = 2
And so, for all values of a and b where the above relationship holds, f is a continuous function.
For part c, a function is differentiable only at points where it is continuous. That said, but continuity alone is not a sufficient condition for differentiability.
Mathematically, a function f is differentiable at a point p if
lim   [f(p+h)-f(p)]/h
h->0
exists.
Technically speaking, a function would be differentiable at a point if the slopes, i.e looking at it from both right and left, both approach this value.
For x < 1, |x-1| = 1-x and so
f(x) = (1-x) + 2 = 3-x
which means that the slope f'(x) = -1
For x > 1
f(x) = ax² + bx
and the slope f'(x) = 2ax + b
For f(x) to be differentiable at x = 1, the slopes have to be equal
2ax + b = -1
2a(1) + b = -1
2a + b = -1
But of course f(x) must be continuous, which takes us back to our continuity condition
a + b = 2
Combining the equations we have
a = -3 and b = 5

Graph f(x) with these values and see what happens at x = 1


You're always welcome.
Regards.