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Simple differentiation question

Hi, so when using differentiation to work out the minima, what’s the point if we can use completing the square to do the exact same thing? Unless there are examples where differentiation would be easier to do.
Original post by subbhy
Hi, so when using differentiation to work out the minima, what’s the point if we can use completing the square to do the exact same thing? Unless there are examples where differentiation would be easier to do.


You are talking about transforming a quadratic equation into a bracket term? as a rule it is easier for you to determine the zeros with it. You can use the bracket term to determine your extreme point for minima.
Reply 2
Original post by subbhy
Hi, so when using differentiation to work out the minima, what’s the point if we can use completing the square to do the exact same thing? Unless there are examples where differentiation would be easier to do.


You can only complete the square for quadratic equations, and even then, if the equation is unpleasant to deal with (e.g: fractional coefficients everywhere), completing the square is going to be a bit clunky to do.

For situations like this, or for a non-quadratic such as a cubic, differentiation is the easiest and simplest method of finding stationary points. (:
Reply 3
You can only complete the square for quadratic equations, and even then, if the equation is unpleasant to deal with (e.g: fractional coefficients everywhere), completing the square is going to be a bit clunky to do.

For situations like this, or for a non-quadratic such as a cubic, differentiation is the easiest and simplest method of finding stationary points. (:


Ah thanks this makes sense!

Also could you help with this: for a cubic, when working out the minima and maxima I’m left with 2 coordinates (-1/3, 1.7) and (2, -11).

Instead of substituting the x value into the second derivative why can’t I look at the y-value and determine If that coordinate represents the minima or maxima. For example, the first one has a value of 1.7 for y so is obviously higher up than -11 and thus the maxima. Not sure if that’s valid.
Reply 4
Original post by subbhy
Ah thanks this makes sense!

Also could you help with this: for a cubic, when working out the minima and maxima I’m left with 2 coordinates (-1/3, 1.7) and (2, -11).

Instead of substituting the x value into the second derivative why can’t I look at the y-value and determine If that coordinate represents the minima or maxima. For example, the first one has a value of 1.7 for y so is obviously higher up than -11 and thus the maxima. Not sure if that’s valid.

For polynomials, yes you can conclude like that. Though it really depends on what your instructor/teacher requires.
I for one never justifies max/min by second derivative test (mostly because I'm bad at differentiation). I look at how signs change for the first derivative. Or if I'm really in a hurry, I'll just quickly sketch the graph of it (because I'm good at curve sketching). But again ask before attempting these "alternative methods".

But beware this logic of "bigger means max, smaller means min" doesn't work for all things. For instance, rational functions like x + 1/[(x+1)(x-1)]. The "higher stationary point" is actually min here.
(edited 11 months ago)
Original post by subbhy
Ah thanks this makes sense!

Also could you help with this: for a cubic, when working out the minima and maxima I’m left with 2 coordinates (-1/3, 1.7) and (2, -11).

Instead of substituting the x value into the second derivative why can’t I look at the y-value and determine If that coordinate represents the minima or maxima. For example, the first one has a value of 1.7 for y so is obviously higher up than -11 and thus the maxima. Not sure if that’s valid.


Just because 1.7 is higher than -11, it is not said that this is the maxima at that y value. It depends on the function itself. In general, it is:

ax³ + bx² + cx + d

The parameters you see are able to move the graph up or down of the axis. That makes it possible that maximums are even at a negative y value. That is why you need to work out the second derivation and to substitue one of the x values in this funciton to make it mathematically clear what kind of extreme point it is.
(edited 11 months ago)
Reply 6
Original post by Kallisto
Just because 1.7 is higher than -11, it is not said that this is the maxima at that y value. It depends on the function itself. In general, it is:

a³x + bx² + cx + d

The parameters you see are able to move the graph up or down of the axis. That makes it possible that maximums are even at a negative y value. That is why you need to work out the second derivation and to substitue one of the x values in this funciton to make it mathematically clear what kind of extreme point it is.


Do you think you could sketch out an example where that isn’t necessarily the case please?
Reply 7
The graph of y = x + x-1 is an example. It has a minimum at y = 2 and a maximum at y = -2 with an asymptote at x = 1.
(edited 11 months ago)
Reply 8
Original post by Kallisto
Just because 1.7 is higher than -11, it is not said that this is the maxima at that y value. It depends on the function itself.
FWIW, I'm pretty sure that for polynomials (or indeed any function with a continuous derivative), the smallest valued critical point is going to be a minima (or point of inflection), the largest a maxima (or point of inflection). In the case of a cubic with 2 turning points we can't have the inflection scenario to worry about either.

That is why you need to work out the second derivation and to substitue one of the x values in this funciton to make it mathematically clear what kind of extreme point it is.What you actually need is for the derivative to change sign at the critical point. It's not that uncommon for it to be easier to determine this directly rather than looking at the 2nd derivative.
Reply 9
Original post by tonyiptony
For polynomials, yes you can conclude like that. Though it really depends on what your instructor/teacher requires.
I for one never justifies max/min by second derivative test (mostly because I'm bad at differentiation). I look at how signs change for the first derivative. Or if I'm really in a hurry, I'll just quickly sketch the graph of it (because I'm good at curve sketching). But again ask before attempting these "alternative methods".

But beware this logic of "bigger means max, smaller means min" doesn't work for all things. For instance, rational functions like x + 1/[(x+1)(x-1)]. The "higher stationary point" is actually min here.


Great thanks!
Reply 10
Original post by Kallisto
Just because 1.7 is higher than -11, it is not said that this is the maxima at that y value. It depends on the function itself. In general, it is:

ax³ + bx² + cx + d

The parameters you see are able to move the graph up or down of the axis. That makes it possible that maximums are even at a negative y value. That is why you need to work out the second derivation and to substitue one of the x values in this funciton to make it mathematically clear what kind of extreme point it is.



Thanks!
Original post by subbhy
Hi, so when using differentiation to work out the minima, what’s the point if we can use completing the square to do the exact same thing? Unless there are examples where differentiation would be easier to do.


Its because not every quadratic function can be easily put into the complete square form.

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