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C3 Differentiation Help Please. ;) watch

1. Hi,

Now I have only done a couple of these so I'm still a bit "crap".

Find the maximum and minimum values of the function f given by,

f(x) = x + 1/(x-2). (x is not = to 2).

Now I started by differentiating the function but straight away I think I have done it wrong.

dy/dx = 1 - (x-2)^-2?? = 1 - 1/(x-2)^2??

I get x = +3 when re-arranged.

Thanks.
2. Yup, your differentiation is correct and there's a turning point at x = 3.

In order to determine maximum + minimum values though, you need to sketch f(x).

*Edit*

There's another turning point also...
3. (Original post by Swayum)
Yup, your differentiation is correct and there's a turning point at x = 3.

In order to determine maximum + minimum values though, you need to sketch f(x).
Oh I think I know what I have done wrong. When I subbed x into y i didn't sub it into the original equation I subbed it into dy/dx. Thats why I was struggling with the first turning point actually. When subbing into original equation I get y = 4.

Now how do I get the other turning point after doing d2y/dx2?? I know I have done this in C1 but it was ages ago.

For d2y/dx2 I get +1/(x-2)^3?? Thanks again!
4. (Original post by FutureMedic)
Oh I think I know what I have done wrong. When I subbed x into y i didn't sub it into the original equation I subbed it into dy/dx. Thats why I was struggling with the first turning point actually. When subbing into original equation I get y = 4.

Now how do I get the other turning point after doing d2y/dx2?? I know I have done this in C1 but it was ages ago.

For d2y/dx2 I get +1/(x-2)^3?? Thanks again!
Just because you know that there's a turning point at (3,4), it doesn't tell you anything about whether it's a maximum or a minimum of a function. Turning points usually only tell you that they're a LOCAL maximum or LOCAL minimum, they're not GLOBAL. The question is asking for global maxima and minima.

In order to find the second turning point though, you don't use d2y/dx^2, you still only use dy/dx = 0. You just missed a solution when solving it - try again.

d2y/dx^2 is used for CLASSIFYING turning points as LOCAL maxima/minima.

Sketch it after you've got the two turning points and classified them.
5. (Original post by FutureMedic)
Oh I think I know what I have done wrong. When I subbed x into y i didn't sub it into the original equation I subbed it into dy/dx. Thats why I was struggling with the first turning point actually. When subbing into original equation I get y = 4.

Now how do I get the other turning point after doing d2y/dx2?? I know I have done this in C1 but it was ages ago.

For d2y/dx2 I get +1/(x-2)^3?? Thanks again!
Just because you know that there's a turning point at (3,4), it doesn't tell you anything about whether it's a maximum or a minimum of a function. Turning points usually only tell you that they're a LOCAL maximum or LOCAL minimum, they're not GLOBAL. The question is asking for global maxima and minima.

In order to find the second turning point though, you don't use d2y/dx^2, you still only use dy/dx = 0. You just missed a solution when solving it - try again.

d2y/dx^2 is used for CLASSIFYING turning points as LOCAL maxima/minima.

Sketch it after you've got the two turning points and classified them.

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