Let $f \in C[a,b]$ be a function whose derivative exists on $(a,b)$. Suppose $f$ is to be evaluated at $x_0$
in $(a,b)$, but instead of computing the actual value $f(x_0)$, the approximate value,
$\hat f(x_0)$, is the actual
value of $f$ at $x_0 + \epsilon $, that is, $\hat f^~(x_0 ) = f(x_0 + \epsilon)$.
Use the Mean Value Theorem to estimate the absolute error $|f(x_0 )−\hat f(x_0 )|$ and the relative
error $|f(x_0 )−\hat f(x_0 )|/|f(x_0)|$, assuming $f(x_0 ) \neq 0$.
Here's what I tried:
Absolute error:
$|f(x_0 )−\hat f(x_0 )|=|f(x_0)-f(x_0+\epsilon)|=|f'(c)|\,\epsilon\,$ where $c \in (x_0, x_0 + \epsilon)$.
And relative error:$$\frac{|f'(c)|\epsilon}{|f(x_0)|}$$
where $c \in (x_0, x_0 + \epsilon)$.
I'm not sure if this is right, and if it is how do I find $c$.
Best Answer
You pretty much have the right idea. As you pointed out, $$\left|f(x_0)-f(x_0+\epsilon)\right|=\left|f^\prime(c)\right| \epsilon$$ for some $c$ in the interval $I = (x_0, x_0+\epsilon)$. Therefore, $$\left|f(x_0)-f(x_0+\epsilon)\right|\leq \epsilon \cdot \sup_{c\in I}\left|f^\prime(c)\right|.$$ Divide both sides by $|f(x_0)|$ to get a bound on the relative error.