Finding turning points on a gradient

In summary, to find the turning points on the curve, we must set the gradient equal to 0 and solve for x. However, we must also consider the possibility of critical points at any x value where the denominator is 0. In this case, the denominator is never 0, so we only need to set the numerator equal to 0 and solve for x, giving us x=±3 as the turning points on the curve.
  • #1
thomas49th
655
0

Homework Statement


The gradient of the curve is:
[tex]\frac{9-x^{2}}{(9+x^{2})^{2}}[/tex]

Find the turning points on the curve

Homework Equations





The Attempt at a Solution



Well for a turning point the gradient of the curve = 0
[tex]\frac{9-x^{2}}{(9+x^{2})^{2}} = 0[/tex]

but now what to do. in the mark scheme it seems that they have multiplied by the denomintoar, but won't that loose roots?

Thanks
 
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  • #2
No. A fraction, a/b is equal to 0 if and only if its numerator is 0. The denominator is irrelevant.

Of course, the fraction is not defined when the denominator is 0 so you might have a critical point (max or min) there, but that is not considered a "turning point". In any case, the denominator here is (9+ x2)2 which is never 0.

By the way, you are right that the gradient being 0 is a necessary condition for a turning point but it is not a sufficient condition. The function f(x)= x3 has f '(0)= 0 but x= 0 is not a turning point. The derivative must change signs at a turning point.
 
  • #3
so what your saying is that i can't get rid of the demoinator

giving me

x = + or - 3?

can i just ask. what is the rules for loosing roots. when can you loose roots?

Thanks
 
  • #4
Your answer is correct. You are not losing roots. As HallsofIvy stated solving for 0 in this case involves only the numerator because a fraction is equal to 0 iff the numerator is equal to 0.
 
  • #5
Here's an example of what it means to "lose roots" Suppose you have the equation f(x)g(x) = f(x)h(x). Note that we can cancel out f(x) on both sides. If we do that, we are left with g(x) = h(x) and we can then solve for x. However, we must also consider the possibility that some value of x exists such that f(x) = 0 and hence also satisfy f(x)g(x)=g(x)h(x). When we cancel out f(x) on both sides, we are effectively throwing away that x value, so we have to separately consider and solve for that.

In this case you would not be discarding roots because there's no way [itex]\frac{1}{(9+x^2)^2}[/itex] can ever be zero for any value of x.
 

1. How do you define a turning point on a gradient?

A turning point on a gradient is a point where the slope of the curve changes from positive to negative or vice versa. This point is also known as a critical point or an inflection point.

2. What is the significance of finding turning points on a gradient?

Finding turning points on a gradient is important in understanding the behavior of a curve and determining its maximum and minimum values. It also helps in identifying the points of change in a system or process.

3. What is the mathematical method for finding turning points on a gradient?

The mathematical method for finding turning points on a gradient is by calculating the derivative of the function and finding the points where the derivative is equal to zero. These points are then analyzed to determine if they are turning points.

4. Can turning points on a gradient occur at other points besides where the derivative is equal to zero?

Yes, turning points on a gradient can occur at other points besides where the derivative is equal to zero. This can happen when the derivative is undefined, or when it changes from positive to negative or vice versa, without passing through zero.

5. How can finding turning points on a gradient be applied in real-world scenarios?

Finding turning points on a gradient has various applications in real-world scenarios, such as in economics, engineering, and physics. For example, it can be used to determine the maximum profit or minimum cost in a business, the optimal design for a structure, or the points of change in a physical system.

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