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    8.
    A curve has the equation y=3x^3 - 7x + \frac{2}{x}.
    a,Verify that the curve has a stationary point when x=1
    b, The tangent to the curve at this statsionary point meets y-axis at the point Q. Find the co-ordinates of Q.

    For, a, i have differentiated to get 9x^2 - 2x^-2 - 7.
    After this, i couldnt factorise it like i normally would so i'm not sure how to work out the stationary points?

    b, i have differentiated again for this as well to get the answer as above^^. But i'm not sure what to to after this either?
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    (Original post by Chelsea12345)
    8.
    A curve has the equation y=3x^3 - 7x + \frac{2}{x}.
    a,Verify that the curve has a stationary point when x=1
    b, The tangent to the curve at this statsionary point meets y-axis at the point Q. Find the co-ordinates of Q.

    For, a, i have differentiated to get 9x^2 - 2x^-2 - 7.
    After this, i couldnt factorise it like i normally would so i'm not sure how to work out the stationary points?

    b, i have differentiated again for this as well to get the answer as above^^. But i'm not sure what to to after this either?
    a) Check the question. You're not asked to find the stationary points, only to verify that x=1 gives a stationary point. So, sub x=1 into dy'dx and confirm that it does indeed equal zero.

    b) Since it's a stationary point, the tangent is horizontal, and you just need to work out the y value when x=1. And hence the co-ordinates of Q.
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    (Original post by Chelsea12345)
    8.
    A curve has the equation y=3x^3 - 7x + \frac{2}{x}.
    a,Verify that the curve has a stationary point when x=1
    b, The tangent to the curve at this statsionary point meets y-axis at the point Q. Find the co-ordinates of Q.

    For, a, i have differentiated to get 9x^2 - 2x^-2 - 7.
    After this, i couldnt factorise it like i normally would so i'm not sure how to work out the stationary points?

    b, i have differentiated again for this as well to get the answer as above^^. But i'm not sure what to to after this either?
    If  x = 1 is a stationary point, then  y'|_{x=1} =0

    b) Note that the gradient is 0 at a stationary point, therefore the tangent at this stationary point will have a gradient of 0. Which means it will be a horizontal line.
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    (Original post by ghostwalker)
    a) Check the question. You're not asked to find the stationary points, only to verify that x=1 gives a stationary point. So, sub x=1 into dy'dx and confirm that it does indeed equal zero.

    b) Since it's a stationary point, the tangent is horizontal, and you just need to work out the y value when x=1. And hence the co-ordinates of Q.
    Yes,you're right! I didn't read the question properly. Thankyou! For the 2nd part,what equation do i have to put x=1 into? The one that i got when i differentiated or the original one? Because if put x=1 into the differentiated equation, i just get 0 again?
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    (Original post by Chelsea12345)
    Yes,you're right! I didn't read the question properly. Thankyou! For the 2nd part,what equation do i have to put x=1 into? The one that i got when i differentiated or the original one? Because if put x=1 into the differentiated equation, i just get 0 again?
    Since you need the y value when x=1, you'd substitute into your original equation y=....
 
 
 
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