Hey guys, need some maths help here! :S

[Muhammad TAimoor]

I'M having problems determining Maxima and Minima of a curve. If anyone feels like helping, sure do help. Paper's on MONDAY!

 

[Binyamine]

I'M having problems determining Maxima and Minima of a curve. If anyone feels like helping, sure do help. Paper's on MONDAY!

There is the first principle, that is first of all you find what is the value of x for which the curve has a maximum or a minimum [stationary point],

take a value less than that x and replace in dy/dx, and a value greater than that x and replace in dy/dx ;


if of this form
Less than x....X..........Greater Than x
(i) +ve 0 -ve this is a maximum
(ii) -ve 0 +ve this is a minimum

Else you could you the second order derivative;
you substitute the value of x for which you had the stationary point in d^2y/dx^2 [Dsquare Y by DxSquare]

if it is > 0 then it is a minimum
if it is < o then it is a maximum
if it is = 0, we cannot use second order derivative, we should use the first principle

 

[Muhammad TAimoor]

There is the first principle, that is first of all you find what is the value of x for which the curve has a maximum or a minimum [stationary point],

take a value less than that x and replace in dy/dx, and a value greater than that x and replace in dy/dx ;


if of this form
Less than x....X..........Greater Than x
(i) +ve 0 -ve this is a maximum
(ii) -ve 0 +ve this is a minimum

Else you could you the second order derivative;
you substitute the value of x for which you had the stationary point in d^2y/dx^2 [Dsquare Y by DxSquare]

if it is > 0 then it is a minimum
if it is < o then it is a maximum
if it is = 0, we cannot use second order derivative, we should use the first principle

Ohkay got it! Thanks alot!

 

[Binyamine]

Ohkay got it! Thanks alot!

Check this ::

An application of second derivative.

 

[irtaza ali]

I'M having problems determining Maxima and Minima of a curve. If anyone feels like helping, sure do help. Paper's on MONDAY!

HAHAHA