Calculate the instantaneous rate of change of (x), given x USING LIMITS

[ClearCCTrue]

Hello, this is a homework question I've attempted several times today, to no avail.
Please note that since I'm currently only in my first unit of calculus, I have not learned the definition of a derivative.
And thus, do not intend to until the next unit. This question is meant to be solved using limits, and that is what I'm trying to do.

The function f(x) = 3/x, where the x = 1/2

This is what I tried:

I assumed the usage of the difference of limit property(Not exactly sure it's called that, nonetheless...)

where , m = lim (f(1/2+h) - f(1/2))
h->0 h

I attempted to sub in: 3/1/2 + h
= 6 + 3/h

f(1/2) = 3/1/2
=

thus: (6 + 3h) - 6 h= 3h/h
= 3

Except the answer at the back of the book is m = -12
Any help would be much appreciated!

 

[stapel]

It's difficult to follow your work, since you used empty-space formatting despite the fact that the forum script strips out those empty spaces (as you saw when you previewed or posted). I think you are attempting to do the following:

. . . . .\(\displaystyle \displaystyle{\lim_{h\, \rightarrow\, 0}\, \frac{f\left(\frac{1}{2}\, +\, h\right)\, -\, f(h)}{h}}\)

However, I can't figure out what you did with f(1/2 + h)...? The substitution should have looked something like:

. . . . .f(1/2 + h) = f((1 + 2h)/2) = 3/[(1 + 2h)/2] = (3/1)*[2/(1 + 2h)] = ...

...and so forth. Then subtract f(h) = 3/h from the result, and divide the whole thing by h.

 

[Quaid]

I assumed the usage of the difference of limit property

Not exactly sure it's called that

The ratio [f(x+h)-f(x)]/h is called a 'difference quotient'.

You're taking a limit of this difference quotient.

I attempted to sub in: 3/(1/2 + h)

= 6 + 3/h

Don't forget to type grouping symbols around denominators when needed (shown in red above). If you omit these grouping symbols, the Order of Operations will not be correct.


Here is your mistake: 3/(1/2 + h) does not simplify to 6 + 3h

It looks like you were thinking that the following is valid.

\(\displaystyle \dfrac{A}{B + C} \; = \; \dfrac{A}{B} + \dfrac{A}{C}\)

Sorry! There is no such rule.


If a sum is in the numerator, instead, then we have this rule.

\(\displaystyle \dfrac{B + C}{A} \; = \; \dfrac{B}{A} + \dfrac{C}{A}\)


So, you have two approaches for simplifying f(1/2 + h).

You can combine 1/2 + h into a single ratio, first. (That's what stapel did.)

Or, you can leave it as 1/2 + h, and then simplify the entire difference quotient.

Cheers :cool:

 

[JeffM]

Hello, this is a homework question I've attempted several times today, to no avail.
Please note that since I'm currently only in my first unit of calculus, I have not learned the definition of a derivative.
And thus, do not intend to until the next unit. This question is meant to be solved using limits, and that is what I'm trying to do.

The function f(x) = 3/x, where the x = 1/2

This is what I tried:

I assumed the usage of the difference of limit property(Not exactly sure it's called that, nonetheless...)

where , m = lim (f(1/2+h) - f(1/2))
h->0 h

I attempted to sub in: 3/1/2 + h
= 6 + 3/h

f(1/2) = 3/1/2
=

thus: (6 + 3h) - 6 h= 3h/h
= 3

Except the answer at the back of the book is m = -12
Any help would be much appreciated!

Like stapel and quaid, I am not sure of what the problem is so any help offered may be off the mark. Use grouping symbols wherever needed under PEMDAS, * for multiplication, / for division, and ^ for exponentiation when posting in the future. Also please read READ BEFORE POSTING.

Like stapel and quaid, I am guessing
that you have been asked to find the slope of the graph of the function \(\displaystyle f(x) = \dfrac{3}{x}\) where \(\displaystyle x = \dfrac{1}{2}.\)

You have been told that the slope of certain functions in x at x = a is shown by the difference quotient:

\(\displaystyle \displaystyle \lim_{h \rightarrow\ 0}\dfrac{f(a + h) - f(a)}{h}.\) How is my guessing so far?

You are QUITE RIGHT that the way to deal with these problems is to apply the various laws of limits. However, in my experience, it usually helps to do as much work as possible BEFORE dealing with limits. So in this specific problem, I would simplify

\(\displaystyle \dfrac{f\left(h + \frac{1}{2}\right) - f\left(\frac{1}{2}\right)}{h}\) before even thinking about limits.

There is no work to be done on the denominator because you cannot simplify h. So work on the numerator.

Given \(\displaystyle f(x) = \dfrac{3}{x},\ f\left(h + \dfrac{1}{2}\right) = \dfrac{3}{h + \frac{1}{2}} = what?\) Just work it out to be as simple as possible.

Now simplify the other term in the numerator: \(\displaystyle f\left(\dfrac{1}{2}\right) = what?\)

Having done that, now subtract them. What do you get?

Now divide that by h, on the assumption that h is not zero. What do you get?

Now you are ready to take limits and use the laws of limits. In other words, you separate the problem into two stages. The first is simplification of the difference quotient. The second is taking limits using the laws of limits. Now the first stage can itself be broken into steps.

The first is to simplify \(\displaystyle f(a + h).\) The second is to simplify \(\displaystyle f(a).\). The third is to compute the difference. The fourth is to divide by h.

A difference quotient is a recipe that tells you exactly what to do.