Calculus HELP!

[Headless_Ninja]

Ok, so I'm gonna look like an idiot for this, but how do I use differentiation to prove that a graph has no turning points?
Help much appreciated!

[The Rose Dragon]

Take the second differential, and if it never equals zero, the graph has no turning points.

I might be wrong, but that's how I remember it.

EDIT: That's wrong. A straight line has no turning points and its second differential always equals zero.

[Headless_Ninja]

Sounds like something I should know. Thanks!

[Hazkali]

Actually, you only need to take the first differential, and show that it cannot be zero in the region that you're interested in. The first differential is zero at all stationary points. The second differential is useful only in working out what sort of stationary point it is.

[randman22222]

...I think you need the first differential, not the second.

[Headless_Ninja]

Ah, OK. But what about if the region I'm interested in is the entire infinite graph?

[randman22222]

Differentiate the equation, and graph the equation. If the graph ever crosses or touches the x-axis, then the original has a turning point.

[Headless_Ninja]

Really? Sounds like a good idea, but I can't rememeber ever having been taught that. Of course, if it works, it works! Thanks!

[BugFix]

Differentiate the equation, and graph the equation. If the graph ever crosses or touches the x-axis, then the original has a turning point.

That works, but unfortunately takes infinite time. I think what you're trying to say is write the derivative as an equation, set it equal to zero, and solve analytically. :)

I actually had to look up "turning point", because that's one of those terms that only has meaning in your calculus textbook and I've long since forgotten it. It happens to be defined as a location where the first derivative is equal to zero, which is the actual jargon people will use in the real world to discuss this stuff, or else they'll replace it will a domain-specific word like "maximum" or "minimum".

[Jayabalard]

hmm... it's been a while.

in the first derivative test, 0s are possible turning points but aren't necessarily turning points.

Looking at the 1st derivative; if the first derivative of a curve is always positive, or always negative, there are no turning points. if it can be both positive and negative then it does have turning points.

[Headless_Ninja]

Well, I think that just about sorts it out. Many thanks to everyone who posted, I thank you from deep within my heart!

[Jayabalard]

hmm, I think I remember the methodology a bit better (though you may have figured it out)


Take the first derivative
Set that equal to 0 and solve for x; this will give you the x location of all possible turning points
for each x in your possible turning point set, check a point on either side and look for differences in sign. Say have 3 possible turning points: x1 = 5, x2= 10, x3= 15. You'd need to compare the signs of

x1- that is < 5 with x1+ that is > 5 but < 10
x2- that is > 5 and < 10 with x2+ that is > 10 but < 15
x3- that is > 10 and < 15 with x3+ that is > 15

[Headless_Ninja]

Yup, sounds familiar. Thanks again!

[averagejoe]

hmm, I think I remember the methodology a bit better (though you may have figured it out)


Take the first derivative
Set that equal to 0 and solve for x; this will give you the x location of all possible turning points
for each x in your possible turning point set, check a point on either side and look for differences in sign. Say have 3 possible turning points: x1 = 5, x2= 10, x3= 15. You'd need to compare the signs of

x1- that is < 5 with x1+ that is > 5 but < 10
x2- that is > 5 and < 10 with x2+ that is > 10 but < 15
x3- that is > 10 and < 15 with x3+ that is > 15



Just remember not to go past any other possible turning point when checking a given turning point when performing step 3. (For example, if x1=5 and x2=5.5, checking x1+1 would be worthless. You'd have to check x1+.25 or something. A minor point, but good to bear in mind.

[Werewindlefr]

Looking at the 1st derivative; if the first derivative of a curve is always positive, or always negative, there are no turning points. if it can be both positive and negative then it does have turning points.
Actually, you can create functions for which this is wrong; although these rarely appear in a math book.

Your definition has an advantage, though, because you can have a turning point without the derivative ever being zero.