As you can see I found the equation but I don't know how to find the points. As far as I tried was $(7, 49)$ but it was wrong.
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2$\begingroup$ What can you say about $S'(x)$ when $(x,x^2)$ is at its closest to $(7,0)$? $\endgroup$ – copper.hat Jul 3 '14 at 2:35
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1$\begingroup$ When you take the derivative and set it equal to $0$, you will get an equation that you will need to solve numerically. $\endgroup$ – André Nicolas Jul 3 '14 at 2:35
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Minimizing the function is the same as minimizing its square. You wrote the equation for the distance, but the exercise asks its square, so we would have $$s(x) = (x - 7)^2 + (y - 0)^2$$ But the point $(x,y)$ is in the parabola $y = x^2$. So we get $$s(x) = (x - 7)^2 + x^4$$
Now, can you solve $s '(x) = 0 $? This will give you the critical points of $s$.
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To find the minimum you need to compute the first derivative of $S(x)$ and find the critical points. We have
$$S'(x)=(4x^3+2x-14)\frac{1}{2\sqrt[]{(x-7)^2+x^4}}=0$$
$$4x^3+2x-14=0$$
This has one real solution: $x\approx1.4087$. See below
http://www.wolframalpha.com/input/?i=zeros+of+4x%5E3%2B2x-14
We know that $S$ must have a local minimum at that point because inspecting the plot of $S'(x)$ shows that to the left of this point the derivative is negative, and to the right the derivative is positive. See below
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I do not know if you heard yet about the solution of cubic polynomials. A classical method is Cardano's which, for the equation $$4x^3+2x-14=0$$ will show that there is only one real root (the two other are complex conjugated) the exact expression being given by $$x=\frac{\sqrt[3]{63+5 \sqrt{159}}}{6^{2/3}}-\frac{1}{\sqrt[3]{6 \left(63+5 \sqrt{159}\right)}} \simeq 1.40872718612$$
answered Jul 3 '14 at 5:49
