I will do the part with all three points and you can do the other with two points.
We are given that $f(x) = e^{2x} - x$, $x_0 = 1$, $x_1 = 1.25$, and $x_2 = 1.6$.
We are asked to construct the interpolation polynomial of degree at most two to approximate $f(1.4)$, and find an error bound for the approximation.
You stated that you know how to find the interpolating polynomial, so we get:
$$P_2(x) = 26.8534 x^2-42.2465 x+21.7821$$
The formula for the error bound is given by:
$$E_n(x) = {f^{n+1}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
Since we do not know where $\xi(x)$ is, we will find each error over the range and multiply those together, so we have:
$$\max_{(x, 1, 1.6)} |f'''(x)| = \max_{(x, 1, 1.6)} 8 e^{2 x} = 196.26$$
Next, we need to find:
$$\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)| = 0.00754624$$
Thus we have an error bound of:
$$E_2(x) = \dfrac{196.26}{6} \times 0.00754624 \le 0.246838$$
If we compute the actual error, we have:
$$\mbox{Actual Error}~ = |f(1.4) - P_2(1.4)| = |15.0446 - 15.2698| = 0.2252$$
Best Answer
I will map it out and you can fill in the details.
To find the derivative, we use:
$$f'(x) = \frac{f(x + h) - f(x - h)}{2h} - \frac{h^2}{6} f^{(3)}(\xi_0)$$
where $\xi_0 \in (x-h, x+h)$.
For your problem:
$$\tag 1 f'(1.2) = \frac{f(x + h) - f(x - h)}{2h} = \frac{f(1.3) - f(1.1)}{2 \times 0.1}$$
The error bound will be given by:
$$\mbox{Max}~ \left|- \frac{h^2}{6} f^{(3)}(\xi_0)\right|, \xi_0 \in (x-h, x+h)$$
The actual error will be given by:
$$|\mbox{Actual value} - \mbox{Calculated value from}~(1)|$$