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How to prove there are no turning points

I need help with this question, prove that there are no turning points on y=e^x+x^3
Differentiate and show that dydx≠0\frac{dy}{dx} \neq 0
Reply 2
Avatar for mdajoka
mdajoka
11
Well just as Mr M pointed out above. If there were any turning points then the derivative at these points would be 0 (that is the tangent to the graph at that point would be a straight horizontal line). So to prove that there are no turning points you have to show that dydx≠0\frac{dy}{dx} \neq 0.
(edited 11 years ago)
Original post by Mr M
Differentiate and show that dydx≠0\frac{dy}{dx} \neq 0


Out of interest, how would you go about finding the turning points with respect to a given angle?

So, taking f(x) = ln(x) as an example, there are no 'normal' turning points, but if you rotate the graph roughly 45* clockwise, then there will be a turning point.

Do you see what I mean?
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by The Polymath
Out of interest, how would you go about finding the turning points with respect to a given angle?

So, taking f(x) = ln(x) as an example, there are no 'normal' turning points, but if you rotate the graph roughly 45* clockwise, then there will be a turning point.

Do you see what I mean?


What an odd question :smile:

I guess that the best approach would be to find a tangent to the original curve with the equivalent angle ... if you have a tangent that hits the x-axis at p degrees then turn p clockwise that tangent is horizontal and give a turning point
Original post by TenOfThem
What an odd question :smile:

I guess that the best approach would be to find a tangent to the original curve with the equivalent angle ... if you have a tangent that hits the x-axis at p degrees then turn p clockwise that tangent is horizontal and give a turning point


If you had, for example, a graph showing the GDP of an economy, and it went sort of like a negative quadratic, but with the 'down-sloping half' as a waving line (like a sin(x) type of graph but at an angle), then it could be interesting to find where the 'relative turning points are', that is, where the line is following the trend growth.

Does that make sense?
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by The Polymath


Does that make sense?


Possibly to an economist :biggrin:

Did my answer make sense?
Original post by TenOfThem
Possibly to an economist :biggrin:

Did my answer make sense?


Yes, it made sense, sort of :smile:

Perhaps I should explain again - if you've got any graph, which goes:

/ for x < 4
_ for x = 4

\ for x > 4, but the \ is a wavy line instead of a straight line

and you wanted to find the points where the 'waves' on that trend line matched the trend

is there some sort of funky calculus you can do. I think you do graph rotation in FP2/3?
Sorry I have the exact same question as the one at the beginning with the turning points etc.
Was wondering if anyone could actually show the working. Struggling to make sense of it.
Original post by ZorroTheN00b
Sorry I have the exact same question as the one at the beginning with the turning points etc.
Was wondering if anyone could actually show the working. Struggling to make sense of it.


This is an old thread, just make a new one if you need help with it.

Stationary points are specified when dydx=0\dfrac{dy}{dx}=0. All you need to do here is show that there aren't any solutions to the equation dydx=0\dfrac{dy}{dx}=0. Consider the fact that ex>0e^x>0 and 3x2≥03x^2 \geq 0 which means that the sum is...?
(edited 6 years ago)

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