If we determinate the second-degree interpolation polynomial that pass through the points $(a,f(a)),(b,f(b))$ and $(c,f(c))$ as:

$$P_2(x)=f(a)+f[a,b](x-a) +f[a,b,c](x-a)(x-b)$$

Where the values of $f[a,b]$ and $f[a,b,c]$ are determinated by the divided differences: $$f[a,b]=\dfrac{f(b)-f(a)}{b-a}$$ and $$f[a,b,c]=\dfrac{\dfrac{f(c)-f(c)}{c-b}-\dfrac{f(b)-f(a)}{b-a}}{c-a}$$

How can I prove that the Brent's method to find the minimun of the fuction is defined as: $$X = b - \dfrac{1(b-a)^{2}[f(b)-f(c)]-(b-c)^{2}[f(b)-f(a)]}{2(b-a)[f(b)-f(c)]-(b-c)[f(b)-f(a)]}$$


I've tried to solve the second-degree interpolation polynomial as it's written below, but I couldn't prove the final statement due to a lack of some kind of math manipulation knowledge. I'm grateful for any help!

If we consider $x-a = x-b = \Delta$,

$$P_{2}(x)=f(a)+f[a,b]\Delta +f[a,b,c]\Delta^{2}$$

$$P_{2}(x)=f(a)+C.\Delta + D.\Delta^{2}\Leftrightarrow ax^{2}+bx+c$$

$$P_{2}'(x)= C + 2D.\Delta \Leftrightarrow 2ax+b$$

$$C + 2D.\Delta = 0 \Leftrightarrow b+2ax = 0$$

$$\Delta =- \dfrac{C}{2D} \Leftrightarrow x =- \dfrac{b}{2a}$$

$$\Delta = x-b$$

$$x-b = - \dfrac{C}{2D}$$

$$x = b - \dfrac{C}{2D}$$

But then, after that, I can't go futher to Brent's definition by considering $ C = f[a,b]$ e $D = f[a,b,c]$, given again below:

$$X = b - \dfrac{1(b-a)^{2}[f(b)-f(c)]-(b-c)^{2}[f(b)-f(a)]}{2(b-a)[f(b)-f(c)]-(b-c)[f(b)-f(a)]}$$


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