Solve f(x) Diff. at x=2: Find a,b,c Values

[Crazy_Diamond]

Homework Statement

The problem is:

f(x) = {ax^2 - b / (x - 2), x < 2
1/4(x^2) - 3c, x >= 2

If f is differentiable at x=2, what are the values of a, b, and c?
In case it is hard to see, I have provided an image

2. Homework Equations
I know that the limit definition of a function needs to be used. However that is as far as I thought.

The Attempt at a Solution

When finding the limit for the left side, I receive:

lim x-> 0, xa(x+h)^2 - 2a(x+h)^2 --ax^3 -ax^2h +bh +2ax^2 / (x^2 - 4x + xh - 2h +4)h
Unfortunately I was not able to factor out the h.

When finding for the right side, I get -x/2 but then I can't find out how to find a, b, or c.

 

[SteamKing]

If you have a calculus problem, don't post it in the Pre-calculus HW forum.

 

[Crazy_Diamond]

Ah sorry about that. Thank you for moving my thread.

 

[RUber]

Before using the definition of the derivative, start with continuity.
The two sides have to agree at the junction. ##f(2^-)=f(2^+)##.

 

[Crazy_Diamond]

Before using the definition of the derivative, start with continuity.
The two sides have to agree at the junction. ##f(2^-)=f(2^+)##.

In that case I would receive for left side:
f(2-) = 4a - b / 2 - 2
= 4a - b, x =/= 2

and for right side:
f(2+) = 1 - 3c

So since the limits on both sides of 2 must be equivalent, I will get:
1 - 3c = 4a - b

But now how can I solve for each variable?

 

[Mark44]

In that case I would receive for left side:
f(2-) = 4a - b / 2 - 2

Use parentheses! What you wrote is this:
##4a - \frac{b}{2} - 2##

= 4a - b, x =/= 2

How do you figure that ##\frac{4a - b}{2 - 2} = 4a - b##?

and for right side:
f(2+) = 1 - 3c

So since the limits on both sides of 2 must be equivalent, I will get:
1 - 3c = 4a - b

But now how can I solve for each variable?

 

[Crazy_Diamond]

Use parentheses! What you wrote is this:
##4a - \frac{b}{2} - 2##
How do you figure that ##\frac{4a - b}{2 - 2} = 4a - b##?

True sorry about that.
I'm not sure what happens when ##\frac{4a - b}{2 - 2}## now as I don't know how to factor out (x-2) from the original ##\frac{4a - b}{x - 2}##

 

[Mark44]

True sorry about that.
I'm not sure what happens when ##\frac{4a - b}{2 - 2}## now as I don't know how to factor out (x-2) from the original ##\frac{4a - b}{x - 2}##

What does this tell you about continuity at x = 2?

 

[RUber]

What would have to be true about ##\frac{ax^2-b}{x-2}## for you to be able to cancel out the x-2? Is this necessary for continuity?

 

[Crazy_Diamond]

What does this tell you about continuity at x = 2?

It's dicontinuous? I'm aware that "2" is actually not the value but 1.999... but I don't know how to express that in a mathematical way.

What would have to be true about ##\frac{ax^2-b}{x-2}## for you to be able to cancel out the x-2? Is this necessary for continuity?

I'm not really sure actually. For to cancel out x - 2 it cannot be equal to 2 which it already is not according to the question right? But I'm not sure what you're implying with it being necessary for continuity.

 

[Mark44]

What you want to have happen is that there is a removable discontinuity at x = 2, a "hole." What values can you specify for a, b, and c, so that this is what happens?

 

[Crazy_Diamond]

What you want to have happen is that there is a removable discontinuity at x = 2, a "hole." What values can you specify for a, b, and c, so that this is what happens?

I don't understand. How can there be a "hole" if ##\frac {1} {4} x^2 - 3c## has a value at 2? Doesn't the question imply that it's a continuous function?

 

[Mark44]

I don't understand. How can there be a "hole" if ##\frac {1} {4} x^2 - 3c## has a value at 2? Doesn't the question imply that it's a continuous function?

I'm talking about the other part - ##\frac{ax^2 - b}{x - 2}##

When x gets close to 2, from the left, the denominator gets close to zero. We don't want the graph of this part of the function to head off to infinity. That would imply a non-removable discontinuity. We won't actually have a hole when x = 2, because the other part of the function formula takes over for x >= 2, but we need to figure out the constants a, b, and c so that the curve is nice and smooth.

 

[RUber]

Say, for example that you had ##\frac{x^2-9}{x-3}## this function is not defined at x=3, but everywhere else equivalent to x+3.
If you had
F(x) =##\frac{x^2-9}{x-3}## for ##x\neq 3## and 6 for x=3, this function would be continuous.

 

[Crazy_Diamond]

I'm talking about the other part - ##\frac{ax^2 - b}{x - 2}##

When x gets close to 2, from the left, the denominator gets close to zero. We don't want the graph of this part of the function to head off to infinity. That would imply a non-removable discontinuity. We won't actually have a hole when x = 2, because the other part of the function formula takes over for x >= 2, but we need to figure out the constants a, b, and c so that the curve is nice and smooth.

Oh I see. So to make it factorable, let b = 4a and that way it becomes:
##\frac{a(x^2 - 4)}{x - 2}##
##a(x+2)##

and now
##a(x + 2) = 1 - 3c##
How can this be solved?

 

[RUber]

Now you can put in x=2 and determine the rule for c in terms of a.

 

[RUber]

Differentiating will lead to the unique answer.

 

[Crazy_Diamond]

Differentiating will lead to the unique answer.

I end up with ##c = \frac{1 - 4a}{3}##
What should I differentiate? I differentiated this to end up with:
##c = \frac{\frac{1 - 4a + h}{3} - \frac{1 - 4a}{3}}{h}##
##c = 1/3##
but plugging that into the former gives me a = 0

 

[RUber]

You need to differentiate the functions that you are trying to equate. The limit from the right and left should be the same for the now continuous function to also be differentiable at 2.
Your reduced functions now should be ##a(x+2) ## and ## \frac 14 x^2 -3c##.
This will help you solve for a, and you already have conditions for b and c in terms of a.

 

[Crazy_Diamond]

You need to differentiate the functions that you are trying to equate. The limit from the right and left should be the same for the now continuous function to also be differentiable at 2.
Your reduced functions now should be ##a(x+2) ## and ## \frac 14 x^2 -3c##.
This will help you solve for a, and you already have conditions for b and c in terms of a.

But how can I solve for a if c is in the reduced function? Since by letting ##f(2^-) = (f(2^+)## will give me ##4a = 1 - 3c##, I still cannot solve for a due to the unknown c.

 

[RUber]

That's why you take the derivative (with respect to x) before you put the 2 into the function. Then your constant c should drop out.

 

[RUber]

You have satisfied continuity with infinite solutions dependent on a. When you satisfy differentiability at 2 also, you will have a unique a, and thus unique b and c.

 

[Crazy_Diamond]

You have satisfied continuity with infinite solutions dependent on a. When you satisfy differentiability at 2 also, you will have a unique a, and thus unique b and c.

Ah I forgot about that. I end up with a and 1/2x. And substituting 2 will give me 1 which is the right answer! Thank you so much for your help. I have just question though. Finding the derivative gives me f'(x) so substituting x = 2 is the solution because doing so gives me the limit as x approaches 2 from right/left side?

 

[RUber]

That's right. For the function to be differentiable at a point, the limit of the function and its derivative (slope) from the right/left must agree at that point.
In this problem you were required to draw upon the meaning of both of those requirements in order to find the solution.