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differentiation

Find the range of values of a for which y = x - a/x has no stationary points.

I differentiated it to give me dy/dx = 1 + a/x2

Surely the value of a would differ depending on the value of x so how can i answer this question

The answer book said a is greater than or equal to 0
Your book is wrong. If you take a = 1, you get y = x - 1/x which has no stationary points.

As you say, dydx=1+ax2\dfrac{dy}{dx} = 1 + \dfrac{a}{x^{2}}. Since we're looking at stationary points, where the gradient is 0, we have that dydx=0\dfrac{dy}{dx} = 0.

So,
1+ax2=01 + \dfrac{a}{x^{2}} = 0. Make a the subject and think about what values it can take.
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Phichi
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Original post by bl64
Find the range of values of a for which y = x - a/x has no stationary points.

I differentiated it to give me dy/dx = 1 + a/x2

Surely the value of a would differ depending on the value of x so how can i answer this question

The answer book said a is greater than or equal to 0


Original post by Erebusaur
Your book is wrong. If you take a = 1, you get y = x - 1/x which has no stationary points.

As you say, dydx=1+ax2\dfrac{dy}{dx} = 1 + \dfrac{a}{x^{2}}. Since we're looking at stationary points, where the gradient is 0, we have that dydx=0\dfrac{dy}{dx} = 0.

So,
1+ax2=01 + \dfrac{a}{x^{2}} = 0. Make a the subject and think about what values it can take.




Indeed dydx=1+ax2\displaystyle \frac{dy}{dx}=1+\frac{a}{x^2}. As dydx=0\displaystyle \frac{dy}{dx}=0

dydx=1+ax2⇒1+ax2=0⇒x2=−a\displaystyle \frac{dy}{dx}=1+\frac{a}{x^2} \Rightarrow 1+\frac{a}{x^2}=0 \Rightarrow x^2=-a

If we consider a=1a=1, as Erebusaur said, we'll have x2=−1x^2=-1 which has no real solutions, and hence when a=1a=1, the curve has no stationary points, so how is the book wrong?

If we look at x2=−ax^2=-a

It's clear that the stationary points should occur at x=−ax=\sqrt{-a}

For there to be no stationary points, there must be no real solutions to x. What a values would correspond to this?

a=0a=0 would yield x=0=0x=\sqrt{0}=0

However, using a=0a=0 would make y=xy=x, which has no stationary points, so a=0a=0 is also another value that a could take for the conditions, easy to miss though.
(edited 9 years ago)

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