Homework Problem on Absolute + Relative Error 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, $\tilde{f}(x_0)$, is the actual value of $f$ at $x_0 + \epsilon$, that is $\tilde{f}(x_0) = f(x_0 + \epsilon)$
a) Use the Mean Value Theorem to estimate the absolute error $|f(x_0) - \tilde{f}(x_0)|$ and the relative error $|f(x_0) - \tilde{f}(x_0)|/|f(x_0)|$, assuming that $f(x_0) \neq 0$.
b) If $\epsilon = 5 \cdot 10^{-6}$ and $x_0 = 1$, find bounds for the absolute and relative errors for
i. $f(x) = e^x$
ii. $f(x) = \sin x$
c) Repeat part (b) with $\epsilon = (5 \cdot 10^{-6}) x_0$ and $x_0 =10$
My work
Part a
There is some $c \in [x_0, x_0 + \epsilon]$ where $f'(c) = \frac{f(x_0 + \epsilon) - f(x_0)}{\epsilon} = \frac{\tilde{f}(x_0) - f(x_0)}{\epsilon}$
Absolute Error: $|f(x_0) - \tilde{f}(x_0)| = |f'(c) \cdot \epsilon|$
Relative Error $\frac{|f(x_0) - \tilde{f}(x_0)|}{|f(x_0)|} = \frac{|f'(c) \cdot \epsilon|}{|f(x_0)|}$
(Is this right? I'm suspicious this isn't the answer expected.)
Part b
Here, I simply calculate the absolute/relative error directly. I'm suspicious because I'm not using part (a) and I'm not calculating "bounds", I'm calculating error values.
i. $f(x) = e^x$
Absolute Error: $|f(1 + 5 \cdot 10^{-6}) - f(1)| = |e^{1 + 5 \cdot 10^{-6}} - e| \approx 1.359 \cdot 10^{-5}$
Relative Error: $|e^{1 + 5 \cdot 10^{-6}} - e|/e \approx 5.000 \cdot 10^{-6}$
ii. $f(x) = \sin x$
Absolute Error: $|f(1 + 5 \cdot 10^{-6}) - f(1)| = |\sin (1 + 5 \cdot 10^{-6}) - \sin 1| \approx 2.702 \cdot 10^{-6}$
Relative Error: $\frac{|\sin (1 + 5 \cdot 10^{-6}) - \sin 1|}{\sin 1} \approx 3.210 \cdot 10^{-6}$
Part c
Similar to (b)
i. $f(x) = e^x$
Absolute Error: $|f(10 + 5 \cdot 10^{-5}) - f(10)| = |e^{10 + 5 \cdot 10^{-5}} - e^10| \approx 1.101$
Relative Error: $|e^{10 + 5 \cdot 10^{-5}} - e^{10}|/e^{10} \approx 5.000 \cdot 10^{-5}$
ii. $f(x) = \sin x$
Absolute Error: $|f(10 + 5 \cdot 10^{-5}) - f(10)| = |\sin (10 + 5 \cdot 10^{-5}) - \sin 10| \approx -4.195 \cdot 10^{-5}$
Relative Error: $\frac{|\sin (10 + 5 \cdot 10^{-5}) - \sin 10|}{\sin 10} \approx 4.949 \cdot 10^{-6}$
 A: Since that is the only question you asked, I will focus my answer on it.

(Is this right? I'm suspicious this isn't the answer expected.) 


What you did is correct, but I would write the result slightly different.
\begin{align*}|f(x_0) - \tilde{f}(x_0)| &= |f'(c) \cdot \epsilon| \\ &= |f'(c)|\,|x_0-\tilde{x_0}|\leqslant \max_{c∈(a,b)}|f'(c)|\,|x_0-\tilde{x_0}|.\end{align*}
Now you have an estimate in the form $$\text{"absolute error in the result"} = \text{constant}*\text{"absolute error in the data"},$$
which is better in the sense, that you explicitly see the connection of both errors. Also you don't know the value of $c$, because the Mean Value Theorem only states "There exists a $c$…". So the only thing you can evaluate numerically is $\underset{c}{\max}…$ .
For the relative error we can do the same 
\begin{align*}\frac{|f(x_0) - \tilde{f}(x_0)|}{|f(x_0)|} &= \frac{|f'(c) \cdot \epsilon|}{|f(x_0)|} \\ &= |f'(c)|  \frac{|x_0|}{|f(x_0)|}\frac{|x_0-\tilde{x_0}|}{|x_0|} \\& \leqslant \max_{c∈(a,b)}|f'(c)|  \frac{|x_0|}{|f(x_0)|}\frac{|x_0-\tilde{x_0}|}{|x_0|}.\end{align*}
I think this answer also answers your not asked questions about b) and c).

It is nice to see, that that that factor resembles very much the condition number $$κ(x)=\frac{∂f(x)}{∂x}\frac{x}{f(x)}.$$
In fact, the condition number is a concept that describes how the relative error in the data connects to the relative error in the result (see here.) The reason, why we here get a different result is, that we used the Mean Value Theorem, while the condition number uses Taylor's expansion.
