Given is a function $f(x)=4x^{2}$, which we want to evaluate for $x\in \left [ 1,2 \right ]$, $\widetilde{x}\in \left [ 1,2 \right ]$ is the approximation of $x$.
- What can be the value of the absolute error of $\widetilde{x}$, so the absolute error of $f(\widetilde{x})$ is at most 2?
- What can be the value of the absolute error of $\widetilde{x}$, so the absolute error of $f(\widetilde{x})$ is at most 11%?
Ok, here is my try. I use the following notation: $\bigtriangleup x=x-\widetilde{x} $, $\bigtriangleup f=f(x)-f(\widetilde{x})$. I can use the statement $\bigtriangleup f=\left | f'(x) \right |\bigtriangleup x$. Now i have to look for such $\bigtriangleup x$ that $\bigtriangleup f\leq 2$. Here comes my problem. I relalize that the minimum and the maximum of the function don't lay in the interval $[1,2]$. I see that from $f''(x)=0$ i get $x=0$, which makes me think that this approach can be wrong, because then i can't find $\bigtriangleup x$...
Can anybody help me with this problem, please? I appreciate any ideas and comments. Thank you in advance!
Given that you are approximating over $[1,2]$, you just need to find the maximum of the first derivatve of $f(x)$ over that interval. Let $d=\max_{x \in [1,2]} |f'(x)|$. Can you find that? Then $\Delta f(x) \lt d \Delta x,\quad \Delta x = \frac {\Delta f(x)}d$