# Finding maximum using elementary calculus. (Using Derivatives)

I'm reading these set of online notes and it reads as following:

1. $f(x)- P(x)$=$\frac{(x-x_0)(x-x_1)(x-x_2)}{3!}$$f'''c(x) where c(x) is some point between the minimum and maximum of the points in {x, x_0, x_1, x_2}. 2. Then they say: Let x_1=x_0+h, x_2=x_1+h. 3. Denote \phi_2(x)=(x-x_0)(x-x_1)(x-x_2) 4. We must compute:If we want a uniform bound for x_0 ≤ x ≤ x_2, we must compute \max\limits_{x_0\leq x\leq x_2} |\phi_2(x)| = \max\limits_{x_0\leq x\leq x_2} |(x − x_0) (x − x_1) (x − x_2)| . 1. Using Calculus: \max\limits_{x_0\leq x\leq x_2} |\phi_2(x)|=\frac{(2h^3)}{3\sqrt3} at x=x_1 \pm$$\frac{(h)}{\sqrt3}$ .

I'm confused by Step 5. How did they determine that the maximum was at x=x_1 \pm$$\frac{(h)}{\sqrt3} . I tried setting the derivative equal to 0 but I can't seem to get the same result. The confusing thing is that you can write x_0,x_1, x_2 in terms of h. So I'm not sure if they converted \phi_2(x) in terms of x_1's and h's. Any help would be much appreciated. Sorry for the basic problem, I just really need to know how they derived step 5. Thank you very much. ## 1 Answer Take the derivative of the expression (x-x_0)(x-x_1)(x-x_2), and equate it to 0.$$(x-x_0)(x-x_1)+(x-x_0)(x-x_2)+(x-x_1)(x-x_2)=0$$We know that x_0=x_1-h and x_2=x_1+h, so we have$$\begin{align}[(x-x_1)+h](x-x_1)+[(x-x_1)+h][(x-x_1)-h]+(x-x_1)[(x-x_1)-h]=0\\(x-x_1)^2+h(x-x_1)+(x-x_1)^2-h^2+(x-x_1)^2-h(x-x_1)=0\\3(x-x_1)^2-h^2=0\\(x-x_1)^2=\frac{h^2}{3}\\x=x_1\pm\frac{h}{\sqrt 3}\end{align}\$

• I think the negatives are a bit off the first expression should be x-x1+h not (x-x1)-h – rain Jun 27 '18 at 1:15
• Correct. Just happens that I have both positive and negative terms, so since I've missed all terms, I get the correct answer. Will fix it anyway. Thanks – Andrei Jun 27 '18 at 1:20