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    Dear reader,
    I require help with a question and you seem like the right person for this.
    Find the turning point of the following functions:
    1) y = 0.5x^2 - 2x
    I first found dy/dx: x - 2 = 0 thus x = 2.
    I substituted this into the original equation to find y: 0.5(4) -2(2) = 0 so my
    coordinates are (2,0)
    I found d2y/dx^2 to see if it is a maximum or minimum point: x = 0 which is a
    positive value so I put down it is a minimum point. However, the correct
    answer is that there are NO turning points for the above equation. Please
    enlighten me ASAP. This is driving me nuts. Cheers!
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    (Original post by Wolfram Alpha)
    Dear reader,
    I require help with a question and you seem like the right person for this.
    Find the turning point of the following functions:
    1) y = 0.5x^2 - 2x
    I first found dy/dx: x - 2 = 0 thus x = 2.
    I substituted this into the original equation to find y: 0.5(4) -2(2) = 0 so my
    coordinates are (2,0)
    I found d2y/dx^2 to see if it is a maximum or minimum point: x = 0 which is a
    positive value so I put down it is a minimum point. However, the correct
    answer is that there are NO turning points for the above equation. Please
    enlighten me ASAP. This is driving me nuts. Cheers!
    f(x) = (1/2)x^2 - 2x

f'(x) = x - 2

f''(x) = 1

    When f'(x) = 0, x = 2, so y = (1/2)(4) - 2(2) = -2, so the turning point is at (2, -2).

    f''(2) = 1, since f''(x) is a constant, i.e. does not depend on x. As f''(x) > 0, this is a minimum point.

    This can clearly be seen by plotting the graph (try desmos.com). Either you are looking at the wrong answer, the answer is wrong, or you have missed a range of values of x specified in the question (e.g. "Find the turning point of the following function, for 0<x<1", which would give no solutions).
    • Thread Starter
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    (Original post by ombtom)
    f(x) = (1/2)x^2 - 2x

f'(x) = x - 2

f''(x) = 1

    When f'(x) = 0, x = 2, so y = (1/2)(4) - 2(2) = -2, so the turning point is at (2, -2).

    f''(2) = 1, since f''(x) is a constant, i.e. does not depend on x. As f''(x) > 0, this is a minimum point.

    This can clearly be seen by plotting the graph (try desmos.com). Either you are looking at the wrong answer, the answer is wrong, or you have missed a range of values of x specified in the question (e.g. "Find the turning point of the following function, for 0<x<1", which would give no solutions).
    Thanks very much.
 
 
 
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