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    (Original post by UCASLord)
    Find the range of the below equation, also indicating whether it is a one-to-one function or a one-to-many function.

    f(x) = x^2 - 7x + 10, x ∈ R

    I differentiated:

    dy/dx = x - 7
    x = 7
    d^2y/dx^2 = 1

    So it's a minimum.

    But it's not actually a minimum. I noticed that the point where x = 7 is at y = 10, the same if x = 0. Now, if I take the point between these two coordinates, I get 3 1/2, which is the actual location of the minimum.

    My question is why doesn't this differentiation give me the minimum point?
    \displaystyle \frac{\mathrm d}{\mathrm dx}(x^2) = x?
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    (Original post by UCASLord)
    Find the range of the below equation, also indicating whether it is a one-to-one function or a one-to-many function.

    f(x) = x^2 - 7x + 10, x ∈ R

    I differentiated:

    dy/dx = x - 7
    x = 7
    d^2y/dx^2 = 1

    So it's a minimum.

    But it's not actually a minimum. I noticed that the point where x = 7 is at y = 10, the same if x = 0. Now, if I take the point between these two coordinates, I get 3 1/2, which is the actual location of the minimum.

    My question is why doesn't this differentiation give me the minimum point?
    I guess what _gcx is trying to point out is that  \displaystyle \frac{d}{dx}(x^2) \neq x
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    (Original post by UCASLord)
    Quick question though: is it a complete coincidence that I got the next point where y = 10, or is there a reason behind it?
    Huh? What do you mean?

    \frac{d}{dx}(x^2-7x+10)=2x-7 as other have pointed out. So setting this =0 gives x=\frac{7}{2}=3.5 which is where the min point is. You get y=10 at the y-intercept, which is nothing to do with it. It's a parabola so every other y coordinate point except the minimum will have two distinct x coordinates to go with it. In this case, x=0 and x=7.
 
 
 
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