determine the number of turning points the curve has...

bumblebee123

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Jan 3, 2018
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I don't think I properly understand this question. Would anyone be able to help?

question: for a curve dy/dx = 3x^2 + 75 Determine how many turning points the curve has

I thought that the turning points would be when the gradient = 0, so 0 = 3x^2 + 75 this means, to solve for x, 3( x^2 + 25 ) however this would give me x = sqrt -25

but negative numbers cannot be square rooted. This means x = 0.

the answer says that it has 0 turning points. Why is this?

any help would be really appreciated :)
 

pka

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Jan 29, 2005
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question: for a curve dy/dx = 3x^2 + 75 Determine how many turning points the curve has
I thought that the turning points would be when the gradient = 0, so 0 = 3x^2 + 75 this means, to solve for x, 3( x^2 + 25 ) however this would give me x = sqrt -25
but negative numbers cannot be square rooted. This means x = 0.
the answer says that it has 0 turning points. Why is this?
I do not know how the term "turning point" is being used here. But look at the plot.
 

Dr.Peterson

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Joined
Nov 12, 2017
Messages
4,110
I don't think I properly understand this question. Would anyone be able to help?

question: for a curve dy/dx = 3x^2 + 75 Determine how many turning points the curve has

I thought that the turning points would be when the gradient = 0, so 0 = 3x^2 + 75 this means, to solve for x, 3( x^2 + 25 ) however this would give me x = sqrt -25

but negative numbers cannot be square rooted. This means x = 0.

the answer says that it has 0 turning points. Why is this?

any help would be really appreciated :)
A turning point is a point where the gradient changes sign; it can happen only when the gradient is zero. You're right.

They almost had me fooled, by giving the gradient rather than y itself. They didn't fool you at that point.

But you found that the gradient is never zero, because the equation has no real solutions. Why do you think you can just say that x = 0?? When x = 0, y' = 75, not 0.

When there is no solution, there is no solution. It doesn't mean that 0 is a solution.
 

Jomo

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Dec 30, 2014
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I agree that x=0 is not a point where the gradient is 0. But let us suppose for a moment that when x=0 it is true that the gradient is 0. The thing is that there does NOT have to be a turning point where the gradient is 0. PLEASE look at the function f(x) = x3 and its graph. f' (x)= 3x2 which equals 0 when x=0. Now looking at the graph it shows there is no turning point at x=0. Further investigation shows that f " (x) =6x which equals 0 when x=0. Since there is a sign change around 0 for f"(x) we have a point of inflection at x=0. Also note that around 0, f'(x) does not have a sign change at x=0 (so no extreme point there and so no turning point). An x-value that makes f'(x) equal to 0 or undefined is just a possible extreme point or turning point.
 

Harry_the_cat

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Mar 16, 2016
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The term "stationary point" refers to points where the gradient is 0 , ie local max, local min and points of horizontal inflection (PHI).

The term "turning point" refers to stationary points where the gradient changes sign, ie local max and local min points. That is , the curve "turns" at that point.

So all turning points are stationary points, but not all stationary points are turning points.

In the case posted, the gradient (dy/dx) is never 0, so there are no stationary points. This implies there are no turning points.
 

Jomo

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Dec 30, 2014
Messages
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One last comment. If you shifted the graph to the right 3 units (so the gradient still is never 0) would you now say x=0 or x=3? Thinking this way would make me realize that something is very wrong with saying that the gradient is 0 when x=0 (even though the gradientis never 0!)
 
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