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Math Help - Local Extreme Values Question

  1. #1
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    Local Extreme Values Question

    Let y=f(x) be differentiable and suppose that the graph of f does not pass through the origin. The distance D from the origin to a point P(x,f(x)) of the graph is given by

    D=\sqrt{x^2 + [f(x)]^2}

    Show that if D has a local extreme value at c, then the line through (0,0) and (c,f(c)) is perpendicular to the line tangent to the graph of f at (c,f(c))

    Thanks
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  2. #2
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    Quote Originally Posted by sazafraz View Post
    Let y=f(x) be differentiable and suppose that the graph of f does not pass through the origin. The distance D from the origin to a point P(x,f(x)) of the graph is given by

    D=\sqrt{x^2 + [f(x)]^2}

    Show that if D has a local extreme value at c, then the line through (0,0) and (c,f(c)) is perpendicular to the line tangent to the graph of f at (c,f(c))

    Thanks
    Don't bump.

    You should realise that the gradient of the tangent is f'(c) and the gradient of the line is \frac{f(c)}{c}. If they are normal then their product is equal to -1: f'(c) \cdot \frac{f(c)}{c} = -1 \Rightarrow f'(c) \cdot f(c) = -c.

    Use the chain rule to differentiate D:

    \frac{d D}{dx} = \frac{1}{2 \sqrt{x^2 + [f(x)]^2}} \cdot [2x + 2 f(x) f'(x)].

    \frac{d D}{dx} = 0 \Rightarrow 2x + 2 f(x) f'(x) = 0 \Rightarrow f(x) \, f'(x) = -x.

    Therefore ....
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