So I was given the following prompt:
"The function $f$ has derivatives of all orders for all real numbers and $f^4(x)=e^{2x-1}$. If the $3$rd degree Taylor polynomial for $f$ about $x = 0$ is used to approximate $f$ on the interval $[0, 2]$, what is the Lagrange error bound for the maximum error on the interval $[0, 2]$?"
I know that the error bound formula looks something like the following: $R_3(x)\leq|\frac{f(z)}{4!}(x)^4|$ for approximating the error of a third degree Maclaurin series, and I also understand that once I found the $z$ value of the equation I'd just have to plug in $2$ to get my answer, but I'm confused about how I'd go about finding the $f(z)$ value. Any help would be appreciated!
Best Answer
The right formula is $\left|\frac{f^{(4)}(z)}{4!} x^4\right|$ where $z$ lies between $0$ and $x$. You are given that the interval of interest is $[0,2]$.
We just need an upper bound on it. Notice that $f^{(4)}$ and $x^4$ are both nonnegative increasing function, hence you just need to evaluate their value at $2$ to find an upperbound.