Maximum value of the given function

Question

$$ \begin{array}{l}\text { Let } P\left(x, x^{2}\right) \text { be a point on the parabola } y=x^{2} \\ \text { If } f(x)=\sqrt{(x-3)^{2}+\left(x^{2}-2\right)^{2}}-\sqrt{x^{2}+\left(x^{2}-1\right)^{2}} \text { then } \\ \text { maximum value of } f(x) \text { is } \end{array} $$

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Interpreting from the function given two points $A(3,2)$ and $B(0,1)$ and a parabola $y=x^2$ Let a point $P (x,y)$ on the parabola which full fills the above condition we need to calculate the maximum value AP-BP. Points $P,A,B$ are collinear (not sure) I note down the equation of line $AB$ intersect it to the parabola to find the point $P$. But unfortunately could not get the answer? Somebody help me!!

Answer 1

$$f(x)= \sqrt{\left(x^2-2\right)^2+(x-3)^2}-\sqrt{\left(x^2-1\right)^2+x^2}$$ $$f'(x) = \frac{4 x \left(x^2-2\right)+2 (x-3)}{2 \sqrt{\left(x^2-2\right)^2+(x-3)^2}}-\frac{4 \left(x^2-1\right) x+2 x}{2 \sqrt{x^2+\left(x^2-1\right)^2}}$$ which is zero at $x = \frac{1}{6} (1 - \sqrt{37})$ only. And $f''(x = \frac{1}{6} (1 - \sqrt{37})) = -3.23116 $. Hence its a maxima.

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Additional Info, $$f''(x) = -\frac{\left(4 x \left(x^2-2\right)+2 (x-3)\right)^2}{4 \left(\left(x^2-2\right)^2+(x-3)^2\right)^{3/2}}-\frac{\left(4 \left(x^2-1\right) x+2 x\right)^2}{4 \left(x^2+\left(x^2-1\right)^2\right)^{3/2}}+\frac{8 x^2+4 \left(x^2-2\right)+2}{2 \sqrt{\left(x^2-2\right)^2+(x-3)^2}}+\frac{8 x^2+4 \left(x^2-1\right)+2}{2 \sqrt{x^2+\left(x^2-1\right)^2}}$$

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Answer 2

The thought process is right, but there are two points that you'd get when you intersect the line with the parabola. The line is $y = \frac{x}{3} +1$

$$x^2 = \frac{x}{3} + 1$$

The roots of this are $$x = \frac{1}{6}(1 \pm \sqrt{37})$$

Now, you might ask which root gives you the maximum, and which the minimum? It is clear to see that you want to increase that the minimum will occur when the point is in between the two points, and the maximum will occur when the point is on one side of the two points (try taking a few points and see this for yourself)

Hence, the negative $x$ will do the trick

$$x^* = \frac{1}{6}(1-\sqrt{37})$$