I have to estimate the value of $f(x)=4x^2$ for $x\in [1,2]$, and $x$ is unknown. the approximated value for $x$ is $\tilde x$, which is also in $[1,2]$. What is the maximum absolute error of $x$, such that the absolute error of $f$ is at most $2$ ?
Where is my mistake ?
$f(x)\simeq f(\tilde x)+f'(\tilde x)(x-\tilde x)$
the absolute errors are
$\Delta x= |x-\tilde x|$,$\quad$$\Delta f=|f(x)-f(\tilde x)|$
Hence
$\Delta f=|f'(\tilde x)|\Delta x$
Since $\tilde x$ is also in $[1,2]$ and $f$ is monoton increasing on that interval, the maximum value of $|f'(\tilde x)|$ is at $x=2$, which is $|f'(\tilde x)|\big |_{x=2}=8x\big|_{x=2}=16$
Therefore $\Delta x\le\frac18$
But the assistant marked my solution as incorrect.
Can you say, where my mistake is ?
Thanks.
We have $$ \begin{eqnarray} \lvert f(x) - f(\tilde{x}) \rvert = \lvert f(\tilde{x}) + f'(\tilde{x})(x-\tilde{x}) - (f(\tilde{x}) + f'(\tilde{x})(\tilde{x} - \tilde{x})) \rvert = \lvert f'(\tilde{x})\rvert \Delta x \end{eqnarray} $$ Now remember that we're trying to maximize $\Delta x$ in the equation of $$ \Delta f = 2 \approx \lvert f'(\tilde{x}) \rvert \Delta x $$ which in turn means we want to minimize $\lvert f'(\tilde{x})\rvert$ so instead of looking at $x=2$ we should look at $x=1$ which gives us $$ \min_{x \in [1,2]} \lvert f'(\tilde{x}) \rvert = 8 \implies \max_{x \in [1,2]} \Delta x = \frac{1}{4} $$ given the constraint that $2 \approx \lvert f'(\tilde{x}) \rvert \Delta x$