Further curve sketching help.. (1 Viewer)

[redorange]

find any pts of inflexion on the curve y=1 / x²-16

i get this weird answer..... wot do you people get?

 

[icycloud]

Is that 1/(x²-16) or 1/x² - 16?

For the first case:

y = 1/(x²-16) = (x²-16)^(-1)
y' = -2x(x²-16)^-2
y'' = -2(x²-16)^-2 + 8x² (x²-16)^-3

Set y'' = 0 for possible pts of inflection,

y'' = -2(x²-16)^-2 + 8x² (x²-16)^-3 = 0
Thus, -2(x²-16) + 8x² = 0
Yields x² = -16/3

Therefore there are no real solutions.
Thus, y has no pts of inflections.

--

If in fact you meant 1/x² - 16, then

y=1/x² -16 = x^-2 - 16
y'= -2x^-3
y''=6x^-4

y''=0 yields no solutions, thus there are NO pts of inflection.

Conclusion: In both cases, there are no pts of inflection.

 

[redorange]

thanks, It was the first case

one question though..

how did u get from this line:

y'' = -2(x2-16)-2 + 8x2 (x2-16)-3 = 0

to this:

Thus, -2(x2-16) + 4x2 = 0

 

[Riviet]

He multiplied everything by (x²-16)³ and the 4x² is meant to remain as 8x².

 

[redorange]

argh i cant do this one either..

find pts of infl. y= x²/(x²-9)

 

[icycloud]

He multiplied everything by (x2-16)3 and the 4x2 is meant to remain as 8x2.

Oops thanks for pointing that out. It's hard doing maths on the comp .

find pts of infl. y= x^2/(x^2-9)

y=x^2/(x^2-9)
= (x^2-9)/(x^2-9) + 9/(x^2-9)
= 1 + 9/(x^2-9)

y' = -18x (x^2-9)^(-2)
y'' = 72x^2 (x^2-9)^(-3) - 18(x^2-9)^(-2)

Set y'' = 0, we get 72x^2 - 18(x^2-9) = 0
Thus, x^2 = -3, so there are no real solutions for x.

Thus, there are no pts of inflection once again.
Again, you can do this using the graphical method, saves you differentiating things.

 

[redorange]

thanks again!

whats the graphical method? (as u can see i dont like differentiating -.- )


Oh and whats this?

y=x^2/(x^2-9)
= (x^2-9)/(x^2-9) + 9/(x^2-9)
= 1 + 9/(x^2-9)

 

[icycloud]

whats the graphical method? (as u can see i dont like differentiating -.- )

Oh, that's just graphing the function and then noting that concavity only changes over asymptotes, therefore there are no pts of inflexion. However your best bet is to differentiate in an exam situation, otherwise you might not get full marks!

Oh and whats this?

y=x^2/(x^2-9)
= (x^2-9)/(x^2-9) + 9/(x^2-9)
= 1 + 9/(x^2-9)

That's just simplifying the original expression so it's easier to differentiate (no need to use the quotient rule).

I just made the top x^2-9+9, then split it into (x^2-9)/(x^2-9) + 9/(x^2-9). Notice how the first part cancels to 1, so we get 1 + 9/(x^2-9). Does that explain it?

 

[redorange]

I just made the top x^2-9+9, then split it into (x^2-9)/(x^2-9) + 9/(x^2-9). Notice how the first part cancels to 1, so we get 1 + 9/(x^2-9). Does that explain it?

Sorry... not following....

 

[icycloud]

Sorry... not following....

Well, x^2 = x^2 + a - a, right? After all it's just adding something and then subtracting it back.

So we have x^2 = x^2 - 9 + 9, yes?

So y = x^2/(x^2-9) = (x^2 - 9 + 9)/(x^2-9)

And remember that (a+b)/c = a/c + b/c

So in this case a = x^2 - 9, b = + 9 and c = x^2-9 so we get:

y = x^2/(x^2-9) = (x^2 - 9 + 9)/(x^2 - 9)
= (x^2-9)/(x^2-9) + 9/(x^2-9)
= 1 + 9/(x^2-9)

Because (x^2-9)/(x^2-9) = 1