$$f(x) = \log_{10}(x)$$
Error formula:
$$e < max[1.35;1.45] |(x-1.35)(x-1.37)(x-1.40)(x-1.45)| max[1.35;1.45] |\frac{-6/\ln(10) * 1/x^4}{4!}| = \\ max[1.35;1.45] |(x-1.35)(x-1.37)(x-1.40)(x-1.45)|*0.0327$$
The problem is that in that range, the value of $(x-1.35)(x-1.37)(x-1.40)(x-1.45)$ is so small my calculator only displays zeros.
I solved this another way by calculating the derivative and then factoring it, and using the factors on the zeroes. For factoring I used WolframAlpha.https://www.wolframalpha.com/input/?i=4%28x%5E3-4.1775x%5E2%2B5.81575x+-+2.69817%29+factor
The result I got then way $$e < 2.933*10^{-6} *0.0327 = 9.59091*10^-8$$
Is this correct?
Is there any way of solving this without WolframAlpha?
For the four data points, the interpolating polynomial is
$$P_3(x) = 0.416667 x^3 - 1.85 x^2 + 3.03996 x - 1.62717$$
The formula for the error bound is given by:
$$E_n(x) = {f^{(n+1)}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
So, we have
$$E_3(x) = {f^{(4)}(\xi(x)) \over 4!} \times (x-1.35)(x-1.37)(x-1.4)(x-1.45)$$
The fourth derivative of $f(x)$ is
$$f^{(4)} = -\dfrac{6}{x^4 \ln(10)}$$
The error is given by
$$E_3(x) = \max_{1.35 \le x \le 1.45}\left|-\dfrac{6}{24 \ln(10) x^4}\right| \max_{1.35 \le x \le 1.45}\left| (x-1.35)(x-1.37)(x-1.4)(x-1.45)\right|$$
For the first max, it is clear that it is at the left endpoint
$$\max_{1.35 \le x \le 1.45}\left|-\dfrac{6}{24 \ln(10) x^4}\right| \approx 0.0326881 $$
We find the second max at $x\approx1.43287$ as
$$\max_{1.35 \le x \le 1.45}\left| (x-1.35)(x-1.37)(x-1.4)(x-1.45)\right| \approx 2.93358×10^{-6}$$
Note, we use calculus by finding the derivative, setting it to zero and then testing the absolute value at each critical point and the endpoints.
Thus
$$ E_3(x) \le 9.5893156398 \times 10^{-8}$$
If we plot the interpolating polynomial and points, we have
If we overlay the $P_3(x)$ and $\log_{10}(x)$ plots, we have
The error seems to match the results and this makes sense because a log plot looks linear.