Use the following values and four-digit rounding arithmetic to construct a third Lagrange polynomial approximation to $f(1.09)$.
The function being approximated is $f(x) = \log_{10}(\tan(x))$. Use this knowledge to find a bound for the error in the approximation.
$$f(1.00) = 0.1924$$ $$f(1.05) = 0.2414$$ $$f(1.10) = 0.2933$$ $$f(1.15) = 0.3492$$
I have $P_4(x) = 1.467 x^3-4.040 x^2+4.638 x-1.873$ (rounded to four digits)
$P_4(1.09) \approx 0.2823$
$\log(\tan(1.09)) \approx 0.2826$
So error is of order $10^{-4}$
How do I find the error bound?
Is it $$\left|\frac{f^{(4)}(\mathsf{\xi}(1.09))}{4!}(1.09-1.00)(1.09-1.05)(1.09-1.10)(1.09-1.15)\right|$$
where
$f^{(4)}(x) = \dfrac{8\sec^2\left(x\right)\tan^6\left(x\right)-12\sec^4\left(x\right)\tan^4\left(x\right)+16\sec^6\left(x\right)\tan^2\left(x\right)-6\sec^8\left(x\right)}{\ln\left(10\right)\tan^4\left(x\right)}$
Is there a simpler way to determine the error bound?
The range of $\tan(x)$ is not bounded, so I don't see a way to find error bound without taking derivative.
Source: https://www.iop.vast.ac.vn/~nvthanh/cours/numerical_methods/CH4_fb_p1.pdf
Your approximation is intended to be used on the interval $[1.00,1.15]$ so you can use the maximum values of $\tan x$ and $\sec x$ on that interval to give you the error bound.