AQA Core 4 3.04 Finding the Range of Validity

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No matter how many times I learn this I always seem to forget it

dom_the

Legend! Excellent video, explained very simply but gets the point across perfectly!

You must be a very good teacher/lecturer :)

BibtheChib

Woah..This video was really helpful.
Thanks a lot.

vaneezaahmad

What if there is a composite function for example ln(1+x) + cosx, would the range be -1<x<=1?

Theobbr

I understand the method of finding the range of validity however I am confused on what its significance is, I don't understand what it means, and why is there a modulus? can you explain it to me please?

edwarda.

How do you know whether to use the ≤ or < symbol?, in a question I did (4+2x)^1/2 the answer in the book stated lxl ≤ 2

Alec

Why is there not a negative sign when shifting the fraction to the right side during finding validity?

chemical_x

Instead of bringing the 2^1/3 outside of the brackets (which is slightly more complicated), couldn't you just do |-7x| < 2 and then divide through by 7 (mod takes care of the neg.) giving you |x| < 2/7

Elliot

Hi. How do we find valid value of x for
| 1/(3x) | < 1 ? It's confusing because x is apart of the denominator.... So how do I write it in -1< x < 1 form ?

vic

I don't know if this is a correct question to ask but, why is it that for the (1 + x) ^ n expansion, the range of validity if |X| < 1? Why the modulus?

peanutsee

what if im given a question involving both binomial and partial fraction. I have to expand 2 times and then add. but now its asking me how to write down the set of values for which expression is valid.

in this case it was ((1-x)^-1) + ((1+x/2)^-1) - ((1+x/2)^-2) after expressing in partial fractions.
how to do it???? help plz

ritesha

What if the questions ask you to expand sin^-1x and state the range of values of x for which the expansion is valid

thomaskhiu

what if you have a quadratic function let say power half maybe, how will you find the range..or how will you be able to 1 plus or minus something..

tshovujosaphatkalanja

In the last question if you take (1+(1-7x))^ now apply the condition on mod |1-7x|<1 you will not get the same value of x as you getting previously why ?

mohammadriyaz

My teacher literally copied his entire lecture from this video with every SAME example and explained it in the class. 😑

chemical_x

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