Finding turning points on a gradient

• thomas49th

Homework Statement

The gradient of the curve is:
$$\frac{9-x^{2}}{(9+x^{2})^{2}}$$

Find the turning points on the curve

The Attempt at a Solution

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

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

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.

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

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.

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 $\frac{1}{(9+x^2)^2}$ can ever be zero for any value of x.