BEC'S methods questions

[bec]

theres probably a more mathematical way of doing these, but for methods, intuition seems to work fine

not in this case! the answer is 1/6

sorry i should've said, i can use my calculator for it
so i tried a table of values, and i got ~1.6667...if i knew the fraction/decimal conversion (which i'd forgotten), i'd be able to work it out.

is that the only way? or is there a way the cas calc can find limits in exact form by itself?

[Ahmad]

Hint: Multiply by

[Ahmad]

is that the only way? or is there a way the cas calc can find limits in exact form by itself?

Try limit((rt[x+9]-3)/x,x,0)

[bec]

is that the only way? or is there a way the cas calc can find limits in exact form by itself?

Try limit((rt[x+9]-3)/x,x,0)

wow I didn't know there was a limit button, thanks!
haha it wasn't working and i spent 5min trying to figure out what i was keying in wrong - i went into alpha and actually wrote out the letters "rt" not realising it was just a typo for sqrt...

[Ahmad]

Another way:



(Where x = 9)

[bec]

how did you go from the second step to the third?

[Ahmad]

Definition of derivative (from first principles).

[droodles]

subbing in zero gives 1/6

how does the first part in the second line of working become in the third line?

[Glockmeister]

The  numerator

[bec]

Another way:



(Where x = 9)

i'm struggling with this! i can do it the other way without a problem but this method seems better, like it would work for more questions. i get that you're using the first derivative, and i understand the "rule" of first principles but i don't understand:
1. why the x in the denominator in the orignal eqn has been changed to h, when elsewhere in the question you have just subbed in x=9
2. why the h in the denominator wasn't an issue when h approached zero. at risk of sounding stupid, where did it go??

[Neobeo]

how do you find this accurately?
lim  
x-->0

Here's a simple way with minimal algebra. Construct a right-angled triangle with legs and 3. Then square the hypotenuse, at the same time reflecting the leg of length 3 onto the hypotenuse, then squaring that as well. There will then be two parts with areas x and 9 respectively. Rearrange the part with area x to form a long rectangle, with a width of , and a height that approaches 6. Since we can show that the limit equation is just the reciprocal of the height, it is equal to .

Here's a probably less taxing visual aid:

[Mao]

ahmad has changed it to h for convenience

the first principle of differentiation is that


it just happens that this question closely resembes this
we have gotten rid of x in the original expression, instead we'll call it h so that it looks similar to the above equation.

with that in mind, we end up with:



which closely resembles the first prinicple when x=9

if we let

then

it follows that

hence we end up with

[Ahmad]

Another way:



(Where x = 9)

i'm struggling with this! i can do it the other way without a problem but this method seems better, like it would work for more questions. i get that you're using the first derivative, and i understand the "rule" of first principles but i don't understand:
1. why the x in the denominator in the orignal eqn has been changed to h, when elsewhere in the question you have just subbed in x=9
2. why the h in the denominator wasn't an issue when h approached zero. at risk of sounding stupid, where did it go??

Original question:


Now I changed x to h, which is valid since you'd get the same value,



Then I replace 9 by x, this x is unrelated to the first x, i could've used y or z instead.

If I used z, it would've been



Which is the derivative of the function at z = 9.

[bec]

ok that clears things up, that's exactly what i meant was confusing me

i think i understand, but i can't differentiate it because i don't know how...and i typed it into my calc and got something complicated, not ...

[Ahmad]