- Thread starter
- #1

- Thread starter Jayden
- Start date

- Status
- Not open for further replies.

- Thread starter
- #1

- Feb 5, 2012

- 1,621

Hi Jaden,Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.

1. Can someone please explain what is going on in steps, really simply.

2. what is h?

The graph of the quadratic equation \(f(x)=x^2\) is shown in the picture. \(P\) is the point \((1,1)\) and \(Q\) is the point \(\left(1+h,(1+h)^2\right)\) where \(h\) is any real number. According to the picture \(h\) should be a positive number as well. Note that both of these points are on the graph since they satisfy \(f(x)=x^2\).

The gradient of the line that joins the two point \(P\mbox{ and }Q\) is given by, \(\displaystyle\frac{(1+h)^2-1}{(1+h)-1}=\frac{(1+h)^2-1}{h}\). Which is given as the slope of \(PQ\).

Now we shall generalize this for any two points.

Let, \(P\equiv(x_1,f(x_1))\mbox{ and }Q\equiv(x_1+h,f(x_1+h))\). Then the slope(say \(m_{PQ}\)) of the line \(PQ\) will be,

\[m_{PQ}=\frac{f(x_1+h)-f(x_1)}{(x_{1}+h)-x_1}=\frac{f(x_1+h)-f(x_1)}{h}\]

Hope this clarified all your doubts.

- Thread starter
- #3

Thank you very much. Yes it does.

Once you end up with the final equation, what needs to be done with that in order to get the tangaent of the curve at an instant? I know it involves finding the limit h -> 0 for the equation. But it still confuses in terms of how I should continue.

- Feb 5, 2012

- 1,621

Note that \(h\) is the separation between the two x-coordinates of the points \(P\) and \(Q\). When, \(h\rightarrow{0}\) the point \(Q\) will converge to \(P\). Ultimately you will have the tangent line. The gradient of the tangent(we shall denote this by \(f'(x_1)\)) is therefore given by,Thank you very much. Yes it does.

Once you end up with the final equation, what needs to be done with that in order to get the tangaent of the curve at an instant? I know it involves finding the limit h -> 0 for the equation. But it still confuses in terms of how I should continue.

\[f'(x_1)=\displaystyle\lim_{x\rightarrow 0}\frac{f(x_1+h)-f(x_1)}{h}\]

Now consider our function \(f(x)=x^2\). Try to substitute for \(f(x_1+h)\mbox{ and }f(x_1)\) in the above equation.

- Jan 26, 2012

- 890

The limit as \(h \to 0\) gives you the slope \( m \) of the tangent. You have found the slope to the curve at the point \((1,1)\), so the tangent is the line:Thank you very much. Yes it does.

Once you end up with the final equation, what needs to be done with that in order to get the tangaent of the curve at an instant? I know it involves finding the limit h -> 0 for the equation. But it still confuses in terms of how I should continue.

\( y=m x +c\)

and as it passes through \( (1,1) \) you have:

\(1=m + c\)

or

\(c=1-m\)

CB

- Thread starter
- #6

Wait, so say we have f(x) = x^2

let a be the x value of point 1.

P1 = (a, f(a))

P2 = (a + h, f(a + h))

using the formula m = (y2 - y1) / (x2 - x1)

We get:

m = (f(a + h) - f(a)) / ((a + h) - a)

which then becomes:

m = (((a + h)^2) - (a^2)) / ((a + h) - a)

which is then simplified to:

m = (((a + h)^2) - (a^2)) / h

Am I right?

- Feb 5, 2012

- 1,621

Yes, of course you are correct. So the slope of \(P_{1}P_{2}\) is \(\displaystyle m = \frac{(a + h)^2-a^2} {h}\). But this can be further simplified. Try to expand the square in the numerator.Wait, so say we have f(x) = x^2

let a be the x value of point 1.

I hope you meant "Let a be the x value of point P1"

P1 = (a, f(a))

P2 = (a + h, f(a + h))

using the formula m = (y2 - y1) / (x2 - x1)

We get:

m = (f(a + h) - f(a)) / ((a + h) - a)

which then becomes:

m = (((a + h)^2) - (a^2)) / ((a + h) - a)

which is then simplified to:

m = (((a + h)^2) - (a^2)) / h

Am I right?

Last edited:

- Thread starter
- #8

- Feb 5, 2012

- 1,621

Incorrect. The original fraction is, \(\displaystyle m = \frac{(a + h)^2-a^2} {h}\). You have expanded \((a+h)^2\) but what about the \(-a^2\) term?(a^2 + 2ah + h^2) / h

?

- Thread starter
- #10

- Feb 5, 2012

- 1,621

You can simplify further. Try to factor the numerator and cancel what is common to both the numerator and the denominator.woops

2ah + h^2 / h

- Thread starter
- #12

- Admin
- #13

- Jan 26, 2012

- 4,040

Almost.2ah + h^2 / h

h(2a + h) / h

h(2a) / 1

2ah / 1

2ah?

\(\displaystyle \frac{2ah+h^2}{h}\)

\(\displaystyle \frac{h \left(2a+h \right)}{h}\)

\(\displaystyle 2a+h\)

So the h in the numerator and denominator cancel and the above line is what you are left with.

- Apr 3, 2012

- 37

woops

2ah + h^2 / h

You must use grouping symbols around the numerator to be correct:

Otherwise, what you typed is equivalent to:

[tex]2ah \ + \ \dfrac{h^2}{h}.[/tex]

- Thread starter
- #15

- Feb 5, 2012

- 1,621

Hi Jayden,Howcome you cancel the h outside the brackets instead of the other h? Is that because the other h is inside brackets and therefore attached to the 2a?

Brackets are used to specify the order of operations. If you write, \(h(2a+h)\), this means that you have to add \(2a\mbox{ and }h\) first and then multiply it with \(h\). You can consider, the things which are inside the parenthesis as one unit. Now you have, \(\dfrac{h(2a+h)}{h}\), you have to multiply \(h\) with \(2a+h\) and divide the answer by \(h\). Multiplication and division have the same order of precedence and therefore you can divide the \(h\) in the numerator by the one in the denominator.

To refresh yourself about the order of operations I kindly suggest you to read, this and this.

Kind Regards,

Sudharaka.

- Thread starter
- #17

I have a question. This one always gets me.

Say we have \(\dfrac{9x^2 + 5}{3x^2}\), does that make it \(\dfrac{6x^2 + 5}{1}\)

Also if we have \(\dfrac{6x^2 + 5}{18x^2}\), what would be the simplified answer? This got me in my last test.

EDIT: uhh how do you display fractions in that font?

- Admin
- #18

- Jan 26, 2012

- 4,040

You might need some time to learn the Latex syntax. We have a forum here with good information on how to get started. Also, this is a new topic so please start a new thread. We are more than happy to help you and very happy you are here!

- Status
- Not open for further replies.