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Differentiation, turning points q

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    11i) Differentiate y = x^3 + 3x and what does this tell you about the number of turning points on the curve?

    I've found dy/dx to be 3x^2 + 3 and the answer is that there are no turning points, I don't understand how the dy/dx shows that

    any help would be great, thanks!
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    what value does dy/dx have at a turning point?

    can you make 3x^2+3= that value??
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    (Original post by Magenta96)
    11i) Differentiate y = x^3 + 3x and what does this tell you about the number of turning points on the curve?

    I've found dy/dx to be 3x^2 + 3 and the answer is that there are no turning points, I don't understand how the dy/dx shows that

    any help would be great, thanks!
    dy/dx means the rate of change of y with respect to x, i.e. how much y changes if you change x.
    So it's the gradient/slope of the curve.

    So if dy/dx = 4, then for every increase in x of 1, y will increase by 4. Basically a ratio of y:x, although someone will strike me down for saying it!

    When a curve is at its minima/maxima, it's flat, i.e. the slope is 0, and so dy/dx = 0

    Since you showed that 3x^2 + 3 = dy/dx, then for there to be a turning point, 3x^2 + 3 = 0

    Try solving that equation and you'll realise that you need to do root(-1), which provides no solutions, ergo there is no minima/maxima.
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    (Original post by The Polymath)
    dy/dx means the rate of change of y with respect to x, i.e. how much y changes if you change x.
    So it's the gradient/slope of the curve.

    So if dy/dx = 4, then for every increase in x of 1, y will increase by 4. Basically a ratio of y:x, although someone will strike me down for saying it!

    When a curve is at its minima/maxima, it's flat, i.e. the slope is 0, and so dy/dx = 0

    Since you showed that 3x^2 + 3 = dy/dx, then for there to be a turning point, 3x^2 + 3 = 0

    Try solving that equation and you'll realise that you need to do root(-1), which provides no solutions, ergo there is no minima/maxima.
    Oh okay thanks!

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