HSC Stuff > HSC Mathematics Extension 1
Help with Derivatives
Dragomistress:
How do I do a and e?
I am rather confused as to how to set these questions out, any help would be greatly appreciated.
Thank you!
jamonwindeyer:
--- Quote from: JKFlacka on February 11, 2017, 08:07:47 am ---How do I do a and e?
I am rather confused as to how to set these questions out, any help would be greatly appreciated.
Thank you!
--- End quote ---
Hey mate! Lemme show you:
So that is the process, remember that \(f(x+h)\) is just whatever function we are talking about, with \((x+h\) instead of \(x\) wherever it appears. It will always work out that some terms up the top will cancel, then you'll be able to divide through by \(h\) as I did in the second last step ;D
E is the same but more complicated:
See if you can expand and simplify that top line and take it from there, you should get \(2x+3\), if it doesn't work out I'll finish it for you ;D
Dragomistress:
Just curious, on the Cambridge textbook examples, it uses a statement: "h (is not equal to) 0" before the final answer.
When am I supposed to use it?
Drewballs:
That is because, if you subbed in h in the previous step, the answer would be undefined, therefore you have to simply the fraction to remove the h from the bottom before subbing in h=0
jamonwindeyer:
--- Quote from: JKFlacka on February 11, 2017, 10:05:26 am ---Just curious, on the Cambridge textbook examples, it uses a statement: "h (is not equal to) 0" before the final answer.
When am I supposed to use it?
--- End quote ---
The explanation for the \(h\neq0\) statement appearing where it does is that you've divided through by \(h\) from the previous step. You wouldn't be allowed to do this after evaluating the limit, because as soon as you substitute zero, it breaks (as Drew mentioned). You need to divide through by \(h\) first and the statement justifies that.
To be honest I think it is a little excessive. You don't need it, you'd get full marks in this sort of question without writing that - I've never seen it in a BOSTES sample solution ;D
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