How to simplify this further?

OP

Original post by notnek

Can you post the initial question?

hello there,
The initial question was

Find the equation of the tangent to the curve y= 2^x +2^-x at the point (2, 17/4).

thanku for all help in advance! :smile:

Reply 4

What did they get as a result? I don't see how factorizing can "simplify" an equation in the first place.

A move from

4y = 15\ln 2 (x) - 15 \ln 2 (2) + 17

to

4y = 15\ln 2 (x - 2) + 17

Is perfectly valid...

Reply 5

OP

Original post by aznkid66

What did they get as a result? I don't see how factorizing can "simplify" an equation in the first place.

A move from

4y = 15\ln 2 (x) - 15 \ln 2 (2) + 17

to

4y = 15\ln 2 (x - 2) + 17

Is perfectly valid...

Hello , when determining the nature of any stationary points , if the second derivative is equal to 0, we differentiate again, right? and then for it to be a point of inflection what must the third differential equal to? thanks

Reply 6

The third derivative must exist and not be 0, for this implies that there is a sign change at the second derivative.

Reply 7

Indeterminate

3

Original post by laurawoods

Hello , when determining the nature of any stationary points , if the second derivative is equal to 0, we differentiate again, right? and then for it to be a point of inflection what must the third differential equal to? thanks

For a point of inflection, the third derivative must be non-zero.

Reply 8

davros

16

Original post by aznkid66

The third derivative must exist and not be 0, for this implies that there is a sign change at the second derivative.

Where do you keep getting this "third derivative" thing from? It's got nothing to do with a point of inflexion.

If the 2nd derivative is 0 then you may or may not have a point of inflexion. If the 2nd derivative is -ve on one side of the point and +ve on the other, then that point is a "point of inflexion" (the curvature changes sign).

(The first derivative can be 0 at a point of inflexion, but it MUST NOT change sign at that point - if it does, then you have a max or min.)

Reply 9

davros

16

Original post by Indeterminate

For a point of inflection, the third derivative must be non-zero.

Er, y=x^5, x^7, x^9 etc all have points of inflexion at x=0. The third derivative is irrelevant.

Reply 10

OP

Original post by aznkid66

The third derivative must exist and not be 0, for this implies that there is a sign change at the second derivative.

Hello there, thanks !

Pls can u help me with understading this:
why cant 4ln(2) - 2 be written as ln(2)^4 - 2 ?
when i type into calc they are giving different values..
:smile:

thanks, laurawoods.

Reply 11

Make sure you are doing ln(2^4) and not [ln(2)]^4

Reply 12

OP

Original post by aznkid66

Make sure you are doing ln(2^4) and not [ln(2)]^4

hello there, thanks ...i didnt know it had to be inside...so it is it the number inside raised to k , and not the whole thing being raised to k , right?

thanks :smile:

Reply 13