Left/Right Limits and Derivatives: piecewise function -2x^2+4x for x<0, 4x^2-5 for x>=0

[calculusderogatory]

Hello. I'm struggling with the first part of this question. I thought this would be the calculation of the limit... I would really appreciate the help and how of how to do this?

 

[stapel]

Hello. I'm struggling with the first part of this question. I thought this would be the calculation of the limit... I would really appreciate the help and how of how to do this?View attachment 36426

How did you get that the limit, as [imath]x[/imath] approaches zero from the left, of [imath]-2x^2 + 4x[/imath], is [imath]-4x + 4[/imath]?

Instead, try finding the numerical value of [imath]-2x^2 + 4x[/imath] if [imath]x[/imath] were zero.

Same question for the right-hand limit.

 

[Dr.Peterson]

Hello. I'm struggling with the first part of this question. I thought this would be the calculation of the limit... I would really appreciate the help and how of how to do this?View attachment 36426

The problem itself is rather confusing, especially out of context. Telling us your reasoning would help us a lot, both to understand what it is about, and to see if you are thinking wrong about even parts it seems to say are right.

As I understand it, it is asking you to apply the definition of the derivative to this function, and state both what one-sided limits are to be taken to find f'(0), and what their values are. I suspect you may be finding the derivatives of each piece of the function at x=0, rather than finding the difference quotient for each. Or perhaps you forgot that f(0) = -5, not 0.

 

[calculusderogatory]

in my understanding, I am finding the derivative equations of the piecewise functions. For the first box (approaching 0 from left), I need to find the derivative of -2x^2+4x. I think using the limit definition this would be -2x-2h+4 but that's not correct according to the program.

The green boxes are marked correct by the program (Edfinity), I can't change them. To answer the reasoning-- I got these answers by following a similar example problem unfortunately, so I really do not know why they are correct.

 

[lev888]

in my understanding, I am finding the derivative equations of the piecewise functions. For the first box (approaching 0 from left), I need to find the derivative of -2x^2+4x. I think using the limit definition this would be -2x-2h+4 but that's not correct according to the program.

The green boxes are marked correct by the program (Edfinity), I can't change them. To answer the reasoning-- I got these answers by following a similar example problem unfortunately, so I really do not know why they are correct.

Please post the limit definition of derivative.

 

[Dr.Peterson]

in my understanding, I am finding the derivative equations of the piecewise functions. For the first box (approaching 0 from left), I need to find the derivative of -2x^2+4x. I think using the limit definition this would be -2x-2h+4 but that's not correct according to the program.

The green boxes are marked correct by the program (Edfinity), I can't change them. To answer the reasoning-- I got these answers by following a similar example problem unfortunately, so I really do not know why they are correct.

Please show the details of your work, for all parts. We need to see that in order to help.

 

[Steven G]

You can't talk about the derivative since the function is not continuos. However, you can talk about the left hand and right hand derivatives.

 

[Dr.Peterson]

in my understanding, I am finding the derivative equations of the piecewise functions. For the first box (approaching 0 from left), I need to find the derivative of -2x^2+4x. I think using the limit definition this would be -2x-2h+4 but that's not correct according to the program.

Looking back at this, I realize I was a little hasty this morning; you did show enough of your thinking that I can say something specific.

The left derivative at 0 would be the limit of [imath]\dfrac{f(0+h) - f(0)}{h}[/imath] as h approaches 0 from the left. As I mentioned before, you appear to have forgotten that [imath]f(0) = -5[/imath], not 0.

The quotient you evidently used, which would be appropriate for [imath]x<0[/imath] but not for [imath]x=0[/imath], is [math]\dfrac{(-2(x+h)^2+4(x+h)) - (-2x^2+4x)}{h}=\\\dfrac{(-2x^2-4hx-2h^2+4x+4h) - (-2x^2+4x)}{h}=\\\dfrac{-4hx-2h^2+4h}{h}=\\-4x-2h+4,[/math] which is probably what you meant to type, since it leads to your wrong answer.

Do you see your error yet? Do you see at least that your answer should not have contained [imath]x[/imath], because they asked for the derivative at a specific value of [imath]x[/imath]?

You can't talk about the derivative since the function is not continuous. However, you can talk about the left hand and right hand derivatives.

Actually, you can ask about the derivative, even when it doesn't exist; and then the answer (to part b) is DNE -- which is meant to teach the lesson you are pointing out.

 

[pka]

Hello. I'm struggling with the first part of this question. I thought this would be the calculation of the limit... I would really appreciate the help and how of how to do this?View attachment 36426

[imath]\Large f(x) = \left\{ \begin{gathered} - 2{x^2} + 4x\quad x < 0 \\ 4{x^2} - 5\quad \quad x \geqslant 0 \\ \end{gathered} \right.[/imath] Given that function,
then [imath]\Large {\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 0\; \ne \;\mathop {\lim }\limits_{x \to {0^ + }} f(x) = - 5}[/imath]
That much alone is sufficient to tell a calculus student that the function [imath]f(x)[/imath] is not continuous at [imath]x=0[/imath],
if not continuous then the derivative DNE.