Trouble simplifying fcn: show that (f(x+h)-f(x))/h <=> -1/(x(x+h)) when f(x) = 1/x

[CGHMaastricht]

Trouble simplifying fcn: show that (f(x+h)-f(x))/h <=> -1/(x(x+h)) when f(x) = 1/x

Hi, I am studying economics mathematics but really struggle with my simplifications, especially when fractions are involved.

I need to show that (f(x+h)-f(x))/h <=> -1/(x(x+h)); when f(x) = 1/x (or x^-1)

So to put that in the function;

Demonstrate:

(((x+h)^-1)-x^-1))/h <=> -1/(x(x+h))

So far I (think) I can get the left hand side to:

1/(h(h+x)) - 1/(hx)

But am stuck as to how to take it further

Any help would be much appreciated!

 

[Deleted member 4993]

Hi, I am studying economics mathematics but really struggle with my simplifications, especially when fractions are involved.

I need to show that (f(x+h)-f(x))/h <=> -1/(x(x+h)); when f(x) = 1/x (or x^-1)

So to put that in the function;

Demonstrate:

(((x+h)^-1)-x^-1))/h <=> -1/(x(x+h))

So far I (think) I can get the left hand side to:

1/(h(h+x)) - 1/(hx)

But am stuck as to how to take it further

Any help would be much appreciated!

Without using calculator, can you calculate the value of:

\(\displaystyle \dfrac{1}{3} - \dfrac{1}{4}\)

Follow exact same method.....

 

[CGHMaastricht]

Without using calculator, can you calculate the value of:

\(\displaystyle \dfrac{1}{3} - \dfrac{1}{4}\)

Follow exact same method.....

Thankyou, but I still can't get it quite right...

I get


1/(h(h+x)) - 1/(hx)

=>

(hx-h(x+h))/(h(x+h)*hx)

=>

(h(x-(x+h)))/(h(x(x+h)))

=> (x-x+h)/(x(x+h))

=> -h/(x(x+h))

But it the numerator is supposed to be -1, not -h - have I missed something?

Thankyou for your reply,

 

[stapel]

I need to show that (f(x+h)-f(x))/h <=> -1/(x(x+h)); when f(x) = 1/x (or x^-1)

...I get


1/(h(h+x)) - 1/(hx)

=>

(hx-h(x+h))/(h(x+h)*hx)

How did you get to this step? Showing all of my work, I get the following:

. . . . .\(\displaystyle f(x\, +\, h)\, -\, f(x)\, =\, \dfrac{1}{x\, +\, h}\, -\, \dfrac{1}{x}\)

. . . . .\(\displaystyle \begin{align}\dfrac{f(x\, +\, h)\, -\, f(x)}{h}\, &=\, \dfrac{\left(\dfrac{1}{x\, +\, h}\, -\, \dfrac{1}{x}\right)}{h}

\\ \\ &=\, \left(\dfrac{1}{x\, +\, h}\, -\, \dfrac{1}{x}\right)\, \left(\dfrac{1}{h}\right)

\\ \\ &=\, \dfrac{1}{h\, (x\, +\, h)}\, -\, \dfrac{1}{x\, h}

\\ \\ &=\, \left(\dfrac{1}{h\, (x\, +\, h)}\right)\,\left(\dfrac{x}{x}\right) -\, \left(\dfrac{1}{x\, h}\right)\, \left(\dfrac{x\, +\, h}{x\, +\, h}\right)

\\ \\ &=\, \dfrac{x}{x\, h\, (x\, +\, h)}\, -\, \dfrac{x\, +\, h}{x\, h\, (x\, +\, h)}

\\ \\ &=\, \dfrac{x\, -\, (x\, +\, h)}{x\, h\, (x\, +\, h)}\end{align}\)

...and so forth. I don't get anything near what you've got. Kindly please reply with clarification. Thank you! ;-)

 

[CGHMaastricht]

How did you get to this step? Showing all of my work, I get the following:

. . . . .\(\displaystyle f(x\, +\, h)\, -\, f(x)\, =\, \dfrac{1}{x\, +\, h}\, -\, \dfrac{1}{x}\)

. . . . .\(\displaystyle \begin{align}\dfrac{f(x\, +\, h)\, -\, f(x)}{h}\, &=\, \dfrac{\left(\dfrac{1}{x\, +\, h}\, -\, \dfrac{1}{x}\right)}{h}

\\ \\ &=\, \left(\dfrac{1}{x\, +\, h}\, -\, \dfrac{1}{x}\right)\, \left(\dfrac{1}{h}\right)

\\ \\ &=\, \dfrac{1}{h\, (x\, +\, h)}\, -\, \dfrac{1}{x\, h}

\\ \\ &=\, \left(\dfrac{1}{h\, (x\, +\, h)}\right)\,\left(\dfrac{x}{x}\right) -\, \left(\dfrac{1}{x\, h}\right)\, \left(\dfrac{x\, +\, h}{x\, +\, h}\right)

\\ \\ &=\, \dfrac{x}{x\, h\, (x\, +\, h)}\, -\, \dfrac{x\, +\, h}{x\, h\, (x\, +\, h)}

\\ \\ &=\, \dfrac{x\, -\, (x\, +\, h)}{x\, h\, (x\, +\, h)}\end{align}\)

...and so forth. I don't get anything near what you've got. Kindly please reply with clarification. Thank you! ;-)

I think it would be a fairly safe assumption that what I did was wrong. The trouble I am facing is that I am doing an economics maths course as part of a degree but it has been year since I studied mathematics in high school, so I really cant remember so many of the little basic tricks and procedures.

Anyway, we are together up until

1/(h(x+h)) - 1/hx

Then I used this rule for subtracting fractions:

a/b - c/d = (ad-bc)/bd

which gave me

h(x-(x+h))/h(x(x+h))

and then I took out the h multiplied on the top and bottom to get

x-(x+h)/x(x+h) => -h/x(x+h)

I must however be at least a little bit on the right track as the denominator is what it should be, just some mistake in my calculations has left the numerator as -h when it should be -1

 

[ksdhart2]

I think it would be a fairly safe assumption that what I did was wrong. The trouble I am facing is that I am doing an economics maths course as part of a degree but it has been year since I studied mathematics in high school, so I really cant remember so many of the little basic tricks and procedures.

Anyway, we are together up until

1/(h(x+h)) - 1/hx

Then I used this rule for subtracting fractions:

a/b - c/d = (ad-bc)/bd

which gave me

h(x-(x+h))/h(x(x+h))

and then I took out the h multiplied on the top and bottom to get

x-(x+h)/x(x+h) => -h/x(x+h)

I must however be at least a little bit on the right track as the denominator is what it should be, just some mistake in my calculations has left the numerator as -h when it should be -1

Okay, so you started with this step:

\(\displaystyle \displaystyle \frac{1}{h\left(x+h\right)}-\frac{1}{hx}\)

And then attempted to use this rule:

\(\displaystyle \displaystyle \frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\)

Where the values of the above variables are:

\(\displaystyle a=1;\:b=h\left(x+h\right);\:c=1;\:d=hx\)

Making the appropriate substitutions leaves:

\(\displaystyle \displaystyle \frac{hx-h\left(x+h\right)}{h\left(x+h\right)\cdot hx}\)

Expanding out terms and simplifying a bit gives:

\(\displaystyle \displaystyle \frac{hx-h\left(x+h\right)}{h\left(x+h\right)\cdot hx}=\frac{hx-hx-h^2}{h\cdot hx\cdot \left(x+h\right)}=\frac{-h^2}{h^2x\cdot \left(x+h\right)}\)

It's at this point that your error appears to come in. You mention you canceled an h, when in fact you can cancel an h².

 

[CGHMaastricht]

That's exactly it, thank you very much. I guess the biggest lesson to learn for me would be to always expand out all brackets before cancelling out terms, as I think that is how my error crept in. Thanks again.

 

[Steven G]

Yer doin' fine CGH.
Good results: you'll never do this again :cool:

Denis, I am glad that the student will never make that same mistake again. My question is why do you keep making the same mistakes over and over again??

 

[Deleted member 4993]

Denis, I am glad that the student will never make that same mistake again. My question is why do you keep making the same mistakes over and over again??

Never... Denis does not repeat mistakes ... he finds a new one every time he is missing the corner.....

 

[Ishuda]

I think it would be a fairly safe assumption that what I did was wrong. The trouble I am facing is that I am doing an economics maths course as part of a degree but it has been year since I studied mathematics in high school, so I really cant remember so many of the little basic tricks and procedures.

Anyway, we are together up until

1/(h(x+h)) - 1/hx

Then I used this rule for subtracting fractions:

a/b - c/d = (ad-bc)/bd

which gave me

h(x-(x+h))/h(x(x+h))

and then I took out the h multiplied on the top and bottom to get

x-(x+h)/x(x+h) => -h/x(x+h)

I must however be at least a little bit on the right track as the denominator is what it should be, just some mistake in my calculations has left the numerator as -h when it should be -1

Sometimes it is easier to factor out common factors before simplifying, i.e.

1/(h(x+h)) - 1/hx = [1/h] [1/(x+h) - 1/x]

and now do the [1/(x+h) - 1/x]