C3 Differentiation

If you had to work it out at the point (4, 36) then there is no need to sub in the other value, but I can't give a proper answer as I don't know what the question is asking you to do. What is the question?

This was a stupid post, I didn't read the question properly.

(edited 11 years ago)

Reply 2

9^{-1/2}=\frac{1}{3}

not

\pm \frac{1}{3}

Reply 3

OP

Original post by BabyMaths

9^{-1/2}=\frac{1}{3}

not

\pm \frac{1}{3}

Since when? I'm really confused now.

Reply 4

OP

Original post by claret_n_blue

If you had to work it out at the point (4, 36) then there is no need to sub in the other value, but I can't give a proper answer as I don't know what the question is asking you to do. What is the question?

The question and a large part of the solution are attached to the original post.

Reply 5

Original post by Ronove

Since when? I'm really confused now.

Can't blame you there. Plenty of teachers seem to think that

\sqrt{4}=\pm 2

. It doesn't.

Reply 6

OP

Original post by BabyMaths

9^{-1/2}=\frac{1}{3}

not

\pm \frac{1}{3}

Since you said this (which shook my understanding to the core by the way :tongue:

) I found this page http://gmatclub.com/forum/square-root-always-positive-114114.html which has blown my mind. Thanks for the response, haha!

Reply 7

OP

Original post by BabyMaths

Can't blame you there. Plenty of teachers seem to think that

\sqrt{4}=\pm 2

. It doesn't.

Well I am my own teacher, haven't had a maths teacher since doing my GCSEs in 2004! I have no idea how I would ever have found this out without you saying it. :tongue:

Original post by Ronove

Well I am my own teacher, haven't had a maths teacher since doing my GCSEs in 2004! I have no idea how I would ever have found this out without you saying it. :tongue:

To clarify if you are asked to solve

x^2=4

you write down

x = \pm 2

.

If you are asked to evaluate

\sqrt 4

then the answer is

2

.

Reply 9

Original post by BabyMaths

Can't blame you there. Plenty of teachers seem to think that

\sqrt{4}=\pm 2

. It doesn't.

Woah! What?

Reply 10

OP

Original post by Mr M

To clarify if you are asked to solve

x^2=4

you write down

x = \pm 2

.

If you are asked to evaluate

\sqrt 4

then the answer is

2

.

Thank you! :h:

Reply 11

TenOfThem

Original post by claret_n_blue

Woah! What?

What?

Reply 12

Original post by TenOfThem

What?

Why is it just 2?

Or is this just for exams, in real life it still plus or minus 2 isn't it?

Reply 13

Couple of long threads on the maths teachers forum at TES explain why students aren't clear on this.

http://community.tes.co.uk/forums/p/623065/7840374.aspx#7840374

http://community.tes.co.uk/forums/p/535448/7132713.aspx#7132713

Reply 14

TenOfThem

Original post by claret_n_blue

Why is it just 2?

Or is this just for exams, in real life it still plus or minus 2 isn't it?

No

It never has been

-2 squared does still =4

But the square root of 4 is 2

Reply 15

Original post by claret_n_blue

Why is it just 2?

Or is this just for exams, in real life it still plus or minus 2 isn't it?

\pm \sqrt{4}=\pm 2

\sqrt{4}= 2

Reply 16

james22

16

Original post by claret_n_blue

Why is it just 2?

Or is this just for exams, in real life it still plus or minus 2 isn't it?

No. The square root of x is defined to be the POSITIVE number y which satisfies y^2=x

Reply 17