Question

Please help!! How do you answer this problem? The question is attached as an image.

Answer 1

so far i know the two limits are different because when I put in DNE for f'(0) the computer told me that i got it right but i guessed for that lol

Answer 2

confused
are we finding the left left limit and right limit for f(x) or f'(x)

Answer 3

I'm going to assume f(x) I guess

Answer 4

\[\lim_{x \rightarrow 0^-}f(x)=\lim_{x \rightarrow 0^-} (-2x^2+7x)=? \\ \lim_{x \rightarrow 0^+}f(x)=\lim_{x \rightarrow 0^+}(7x^2-5)=?\]

Answer 5

yo know this question is weird
like it has to be continuous and smooth at x=0 for f'(0) to exist
it is hard to tell if they want us just find the limits of f as x approaches 0
or if they want the limits of f' as x approaches 0

Answer 6

\[\lim_{x \rightarrow 0^-}f'(x)=\lim_{x \rightarrow 0^-}(-2x^2+7x)'=? \\ \lim_{x \rightarrow 0^+}f'(x)=\lim_{x \rightarrow 0^+}(7x^2-5)'=?\]
like either set of limits would actually show f'(0) dne in this case

Answer 7

@Kainui
Using esp what set of limits do you think they are talking about

Answer 8

@jim_thompson5910 hey if you don't mind, can you you please help me answer this question? it looks like @freckles is also having trouble with it.

Answer 9

how far did you get with freckles' suggestions?

Answer 10

umm i think i got it wrong for sure. i thought the derivative for the left hand was 7 and the right hand was 0.

Answer 11

so you typed in "7" in the first box and "0" in the second box? and they said both were wrong?

Answer 12

yes

Answer 13

ok try plugging 0 into both pieces of f(x)

NOT f ' (x)

Answer 14

0 and -5 don't work either

Answer 15

are there any example problems given to you in the book? That are similar to this one

Answer 16

I'm curious about the format they want the answer in.

Answer 17

i don't think so. the only examples using right and left hand limits are either with graphs or simples ones which you can easily plug in and find the limits.

Answer 18

the question is from this online hw thing called webwork that we use for calc class

Answer 19

ok, maybe they want you to type in the expressions for each piece (without you evaluating)

Answer 20

wait, like just using the definition of derivative?

Answer 21

what do you mean?

Answer 22

like f'(x) = f(x+h) - f(x) / h or is that completely wrong?

Answer 23

that is the definition of the derivative, yes

Answer 24

but I was thinking of simply you typing in each piece into each box

so you'd have -2x^2 + 7x in the first box for instance

Answer 25

oh lol yeah that didn't work either

Answer 26

hmm what could they possibly want

Answer 27

ok, try typing in the derivative of each piece

Answer 28

i have no idea. i literally attempted this question 24 times putting in different answers every time (and possibly repeating wrong ones because i lost track)

Answer 29

are the derivatives -4x +7 and 14x because i tried that too ?

Answer 30

yeah those are the correct derivatives

Answer 31

I wonder if this will work.

Try typing in `f(x) = 0` for the first box and `f(x) = -5` for the second box

where am I getting this? from the fact that

\[\Large \lim_{x \to 0^{-}} f(x) = 0\]
\[\Large \lim_{x \to 0^{+}} f(x) = -5\]

Answer 32

yeah i tried that before too but it doesn't work :(

Answer 33

you forgot the f(x) = part

Answer 34

do you mean like this because it says f is not defined in this context.

Answer 35

I guess you could try the same with f ' (x). Otherwise, I'm stumped on what they want.

Answer 36

no it says the same thing unfortunately

Answer 37

definitely frustrating. Usually they would give an example answer so we at least know what format to type in

Answer 38

i know. i don't even know what they want. all i know is that the two limits are different since f'(0) does not exist.

Answer 39

yeah the LHL and RHL have to be equal for the limit to exist

Answer 40

yeah

Answer 41

Sorry I'm stumped. I guess you'll have to ask your teacher because it seems like only s/he would know what the system wants.

Answer 42

okay thanks i'll see what he says

Answer 43

@satellite73 hey do you know how to solve this problem?

Answer 44

how about you @radar ?

Answer 45

Alright.. This is pretty long, but try to read it anyway.

I'd prefer to start with an explanation of the problem first.
We have the following function:
$$
\begin{cases}
-2x^2 + 7x & \mbox{for }x < 0 \\
7x^2 -5 & \mbox{for }x \ge 0 \\
\end{cases}
$$And we want to find the derivative at x=0 (if exists ofc) using the derivative definition.
We can define the derivative in two equivalent ways:
$$
f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \equiv \lim_{p \to x} \frac{f(p) - f(x)}{p - x}
$$In order to try and keep this post shorter I will not explain the definition of the derivative. If you do not understand it don't hesitate to ask and I'll do my best to explain =)

We can use both definitions here, but looking at your answers they are looking to something of the form \(\lim_{x \to 0}\) which fits to the second definition for f'(0):
$$
f'(0) = \lim_{p \to 0} \frac{f(p) - f(0)}{p} = \lim_{x \to 0} \frac{f(x) - f(0)}{x}
$$They just named \(p\) as \(x\), but it is essentially the same.
Now, in order for \(f'(0)\) to be defined, the function \(f(x)\) has to be continuous at \(x=0\) and the derivative limit has to exist.

Assuming you are familiar with the background, we can translate those requirements to conditions that will be easy for us to check.
The first requirement, being continuous at \(x=0\), means:
A) The following limit exists \(\lim_{x \to 0} f(x) = f(0) \)
B) In order for that limit to exist, the function has to be defined at f(0) and in some neighborhood (even small one) around it. Our function is defined everywhere, so it isn't a problem.

The second requirement, of the derivative limit to exist, means that both the left-hand limit and the right-hand limit evaluate into the same value.

In fact, we can already tell that \(f'(0)\) doesn't exist from the first requirement because:
$$
\lim_{x \to 0^-} f(x) = -2(0)^2 + 7(0) = 0 \\
f(0) = 7(0)^2 - 5 = -5 \\
\lim_{x \to 0^-} f(x) \neq f(0)
$$Which means that the function is not continuous at \(x=0\)..

However, if we still were to try and evaluate the derivative limit we'd just see that we get a value of an infinite magnitude (\(\pm \infty\)).

So let's try. Right hand limit:
$$
\lim_{x \to 0^+} \frac{f(x) - f(0)}{x} = \lim_{x \to 0^+} \frac{(7x^2 - 5) - (7(0)^2 - 5)}{x} = \\
= \lim_{x \to 0^+} \frac{7x^2}{x} = \lim_{x \to 0^+} 7x = 7 \\
$$Left hand limit:
$$
\lim_{x \to 0^-} \frac{f(x) - f(0)}{x} = \lim_{x \to 0^-} \frac{(-2x^2 + 7x) - (7(0)^2 - 5)}{x} = \\
= \lim_{x \to 0^-} \frac{-2x^2 + 7x + 5}{x} \\
$$But if you notice.. the numerator gets closer and closer to 5 while the denominator gets closer and closer to 0.. so:
$$
\lim_{x \to 0^-} \frac{-2x^2 + 7x + 5}{x} = \frac{5}{0^-} = -\infty
$$
Which also shows us that the function is not continuous at \(x=0\) and therefore that the derivative \(f'(0)\) could not exist.

=======================================

Now, all I wrote so far tries to explain how to answer such a question, but it doesn't tell you what to type into those textboxes of yours and despite being tempted to post it I realized it isn't worth much without an answer.

So I googled.
And among questions of other frustrated students with many answers that the website doesn't like, I've found something interesting:
https://github.com/djun-kim/webwork-open-problem-library-old/blob/master/NationalProblemLibrary/OSU/high_school_apcalc/dchmwk3/prob8.pg
You're looking at the code that generates your question in the website =)
What a great homework system! you can cheat, but you have to learn programming to do so =p

Reading the code you can see that they generate 4 random numbers \(a,b,c,d\).
Then they print the function as:
\begin{cases}
-ax^2 + bx & \mbox{for }x < 0 \\
cx^2 - d & \mbox{for }x \ge 0 \\
\end{cases}
And the interesting part is on the bottom - the answers.
They are expecting the something of the kind:
$$
\lim_{x \to 0^-} \frac{-a x^2 + bx + d}{x}\\
\lim_{x \to 0^+} \frac{c x^2}{x}
$$In our case:
$$
\lim_{x \to 0^-} \frac{-2 x^2 + 7x + 5}{x}\\
\lim_{x \to 0^+} \frac{7 x^2}{x}
$$If you notice, these appear in my explanation above.
They are expecting in the boxes the following textual forms:
first: (-2*x^2 + 7*x + 5)/x
second: 7*x^2/x
third: DNE

This is a stupid question.. You can understand it perfectly, but one sign misplaced, a redundant space or anything like that could make it all fail..
I do not recommend 'peeking' on the answers generally, but sometimes...

Hope it helps

Answer 46

wow thank you so much!!! that was the right answer :) you are a genius.

Answer 47

Glad it worked, you're welcome =)