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# C2 - Applications of Differentiation Question watch

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

What I have done so far:

y = x-a/x
= x-ax^-1
dy/dx = 1+ax^-2
= 1+a/x^2
a/x^2 = -1
a = -(x^2)
a = -x^2
therefore, a<0

However, I checked the answers and my answer was incorrect. The correct answer is: a is greater or equal to zero. So, could anyone please tell me where I went wrong?

Help would really be appeciated.
2. (Original post by TarotOfMagic)
Q: Find the range of values of a for which y=x-a/x has no stationary points.

What I have done so far:

y = x-a/x
= x-ax^-1
dy/dx = 1+ax^-2
= 1+a/x^2
a/x^2 = -1
a = -(x^2)
a = -x^2
therefore, a<0

However, I checked the answers and my answer was incorrect. The correct answer is: a is greater or equal to zero. So, could anyone please tell me where I went wrong?

Help would really be appeciated.
That went wrong
All correct, wrong conclusion.
3. (Original post by Bath_Student)
That went wrong
All correct, wrong conclusion.
Hmm, do you mind telling me the correct conclusion - or maybe a hint?. Because, I really do not understand where I went wrong.
4. (Original post by TarotOfMagic)
a = -x^2
therefore, a<0
You're saying a = (negative) (square number).

Squared numbers are always positive, so no matter what the value of x is. x^2 is positive. So -x^2 is always negative because it is a negative multiplied by something always positive. So a is always negative means that there is a stationary value. You want a = -x^2 to not be true, which means a >0.
5. (Original post by TarotOfMagic)
Hmm, do you mind telling me the correct conclusion - or maybe a hint?. Because, I really do not understand where I went wrong.
For there to be NO stationary points, the derivative must not equal 0. So when solving, think of it as being when does the derivative not equal 0.

As such, you drew the wrong conclusion. It is the opposite.

Make sense?
6. (Original post by Bath_Student)
For there to be NO stationary points, the derivative must not equal 0. So when solving, think of it as being when does the derivative not equal 0.

As such, you drew the wrong conclusion. It is the opposite.

Make sense?
Thanks.

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