piece-wise and differentiation

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whiteti

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Let f(x) = 7x-8 if x<4
20 if x=4
x2+4 if x>4
Use Newtons Difference Quotient to determine whether or not f is differentiable at x=4; if it is not, state the way in which it is non-differentiable (discontinuity, vertical line or corner).

I do not know how to combine the piecewise and the newtons equations... please help.
 
whiteti said:
Let f(x) = 7x-8 if x<4
20 if x=4
x2+4 if x>4
Use Newtons Difference Quotient to determine whether or not f is differentiable at x=4; if it is not, state the way in which it is non-differentiable (discontinuity, vertical line or corner).

I do not know how to combine the piecewise and the newtons equations... please help.
Click to expand...
The derivative is the limit of the difference quotient, correct?

For a limit to exist, the left limit and right limit must both exist and be equal, correct?

Can you calculate the left limit of the difference quotient at x = 4? What is it?

How about the right limit? What is it?

What does that tell you?
 
JeffM said:
The derivative is the limit of the difference quotient, correct?

For a limit to exist, the left limit and right limit must both exist and be equal, correct?

Can you calculate the left limit of the difference quotient at x = 4? What is it?

How about the right limit? What is it?

What does that tell you?
Click to expand...

yes and yes.
right limit is x2+4 = (4)2+4 = 20
Left limit is 7x-8= 7(4)-8=20
F(4) = 20

this tells me that it is continuous at x = 4. But I haven't used Newtons difference quotient.... Still not sure how to combine the concepts of limits and differentiation
 
whiteti said:
yes and yes.
right limit is x2+4 = (4)2+4 = 20
Left limit is 7x-8= 7(4)-8=20
F(4) = 20

this tells me that it is continuous at x = 4. But I haven't used Newtons difference quotient.... Still not sure how to combine the concepts of limits and differentiation
Click to expand...
WHOA I am not asking you to determine the right and left limits of the function, but the right and left limits of the function's difference quotient.
 
To get you started

The difference quotient at x = 4 is \(\displaystyle \dfrac{f(4 + h) - f(4)}{h} = \dfrac{f(4 + h) - 20}{h}.\)

Now suppose that h > 0

What does the difference quotient become?

What's its limit as h stays positive but approaches 0>

Suppose h < 0.

What does the difference quotient become?

What's its limit as h stays negative but approaches 0?

Are the limits equal?
 
I do not know what the difference quotient becomes... or where we go after the second part you provided.
 
and I do not understand when and how the piece wise equations come into play
 
Ive never done precalculus. I understand you cannot teach a whole course, I just need to know the steps on how to complete this problem.
 
whiteti said:
Ive never done precalculus. I understand you cannot teach a whole course, I just need to know the steps on how to complete this problem.
Click to expand...
Step 1

Given the definition of the function what is f(4 + h) if h is positive?

Step 2

What is f(h + 4) - f(4) = f(h + 4) - 20?

Step 3

So what is \(\displaystyle \dfrac{f(h + 4) - f(4)}{h}\)?

Step 4

Calculate the limit as h approaches 0 from above.

Now repeat steps 1 through 4 on the assumption that h is negative.

Are the right and left limits equal?

If so, the limit exists and equals the derivative.

If not, the limit does not exist and thus neither does the derivative.
 
OK. So can you follow my step by step directions? If you are having trouble, please show us how far you have gotten so we can get you unstuck.
 
kind of

lim + = f(4+h)-x^2+4
h
Lim - = f(4+h)-7x-8
h

If i substitute 4 for x and 0 for h i get -16... Im really confused
 
whiteti said:
If i substitute 4 for x and 0 for h i get -16 ... Im really confused
Click to expand...

You may not substitute 0 for h (in those expressions with h as the denominator) because division by zero is not defined.

Have you studied any algebra? :-?
 
whiteti said:
kind of

lim + = f(4+h)-x^2+4
h
Lim - = f(4+h)-7x-8
h

If i substitute 4 for x and 0 for h i get -16... Im really confused
Click to expand...
The way I like to do these problems is to worry about the limiting process as the last step.

To take the right hand limit of the difference quotient, we must restrict h to positive numbers. In other words, we are interested in the region greater than 4. Does that make sense?

We are interested here in the limit at x = 4. So x really drops out of the analysis. I think that may be one source of your confusion with this problem.

According the definition of f(x): \(\displaystyle f(4) = 20.\) With me to here?

Again according to the definitions: \(\displaystyle h > 0\ and\ x = 4 \implies 4 + h > 4 \implies f(4 + h) = (4 + h)^2 + 4 = 16 + 8h + h^2 + 4 = 20 + 8h + h^2.\)

Did you follow that?

So \(\displaystyle f(4 + h) - f(4) = 20 + 8h + h^2 - 20 = 8h + h^2.\)

So \(\displaystyle \dfrac{f(4 + h) - f(4)}{h} = \dfrac{8h + h^2}{h} = 8 + h.\)

I am just working out what the difference quotient is if h > 0 and x = 4.

Now take the limit as positive h approaches 0

\(\displaystyle \displaystyle \lim_{h \rightarrow 0^+}\dfrac{f(4 + h) - f(4)}{h} = \lim_{h \rightarrow 0^+}(8 + h) = 8.\)

Now you do it for the left limit when h < 0.
 
Thankyou so much Jeffm. I have completed it and got the same answer.
my limit from the right is 8 and from the left is 7. These are not differentiable at x=4 because it is neither continuous from the left or right.

Thankyou for being patient!!!!
 
whiteti said:
Thankyou so much Jeffm. I have completed it and got the same answer.
my limit from the right is 8 and from the left is 7. These are not differentiable at x=4 because it is neither continuous from the left or right.

Thankyou for being patient!!!!
Click to expand...
You're welcome. But you are being a bit ambiguous in your language. The function itself is everywhere continuous, even at x = 4. The limits from right and left of the function are 4^2 + 4 = 16 + 4 = 20 and 7 * 4 - 8 = 28 - 8 = 20. The limit of f(x) as x approaches 4 exists and equals f(4) so f(x) is continuous.

What does not exist is the limit of the difference quotient as x approaches 4 because the right and left limits are not equal. So f(x) is not differentiable at x = 4. You need to be clear about the difference between the function and its difference quotient. I think you should say that the derivative of f(x) is not continuous at x = 4 because it does not exist there.

Differentiability implies continuity, but continuity does not imply differentiability. That was the point of this exercise.
 
I think your conclusion is incorrect.

Did you consider graphing the given function?
 
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