Find the largest value for $x_1$ in (0,1) such that $f(0.5)-P_2(0.5) = -0.25$ (interpolation)

I'm not really sure where to go with this problem and I'm hoping you can help.

The problem states:

Let $f(x) = \sqrt{x - x^2}$ and $P_2(x)$ be the interpolation polynomial on $x_0 = 0, x_1$, and $x_2 = 1$. Find the largest value of $x_1$ in $(0,1)$ for which $f(0.5) - P_2(0.5) = -0.25$

What I've done:

So it's obvious that we need the second degree lagrange polynomial here. Further, $f(0.5) = 0.5$.

So solving for the lagrange polynomials:

$L_0(x) = \frac{(x-x_1)(x-1)}{(0-x_1)(0-1)} = \frac{x^2-(1-x_1)x+x_1}{x_1}$ $L_1(x) = \frac{(x-0)(x-1)}{(x_1 - 0)(x_1 -1)} = \frac{x^2-x}{x_1^2 -x_1}$ $L_2(x) = \frac{(x-0)(x-x_1)}{(1-0)(1-x_1)} = \frac{x^2 - x_1x}{1-x_1}$

and we know

$P_2(x) = L_0(x)f(x_0) + L_1(x)f(x_1) + L_2(x)f(x_2)$

So

$P_2(x) = L_0(x)* 0 + L_1(x)f(x_1) + L_2(x)*0$

So

$P_2(x) = \frac{x^2-x}{x_1^2 -x_1}f(x_1)$

But at this point I dont know where to go. We know $P_2(0.5) = .75$ from the equation in the question. But if you try to solve

$.75 = \frac{-.25}{x_1^2 -x_1}f(x_1)$

You get nothing that makes sense.

Can anyone explain to me what I'm doing wrong here? Thank you!

You did a great job! We have

$$f(x_1) = \sqrt{x_1-x_1^2}$$

We can substitute $f(x_1)$ in:

$$.75 = \dfrac{-.25}{x_1^2 -x_1}f(x_1)$$

This yields (simplify):

$$.75 = \dfrac{-.25}{x_1^2 -x_1}\sqrt{x_1-x_1^2} = \dfrac{0.25}{\sqrt{-(x_1-1) x_1}}$$

Solving for $x_1$ yields two roots in our desired range:

$$x_1= 0.127322, x_1 = 0.872678$$

The problem asked for the largest $x_1 \in (0,1)$, so we choose:

$$x_1 = 0.872678$$

Now, if we actually use this $x_1$ in the three points and find the Lagrange Interpolating Polynomial, we arrive at:

$$P_2(x) = -3x(x-1)$$

Lets check several things:

• $P_2(0) = 0 ~\checkmark$
• $P_2(1) = 0 ~\checkmark$
• $P_2\left(\dfrac{1}{2}\right) = -\dfrac{3}{2}\left(\dfrac{1}{2}-1\right) = \dfrac{3}{4} ~\checkmark$
• $f\left(\dfrac{1}{2}\right)-P_2\left(\dfrac{1}{2}\right) = \dfrac{1}{2}-\dfrac{3}{4} = - \dfrac{1}{4}~\checkmark$
• Thanks again for all your help. You've saved me so much time. – John Oct 14 '14 at 1:19