Answer
$2.09999\leq f(0.1) \leq2.1$
Work Step by Step
Step 1. Recall the results from Exercise 68: $min\ f′\leq\frac{f(b)-f(a)}{b-a}\leq max\ f′$
Step 2. Graphing the given function $f′(x)=\frac{1}{1+x^4}$, we can see that it is a decreasing function as shown in the figure (you can also take the derivative of $f′(x)$ and show that $f′′(x)\lt0$ in the interval).
Step 3. We can find $min f′=f′(0.1)=\frac{1}{1+0.1^4}=\frac{1}{1.0001}$ and $max f′=f′(0)=1$
Step 4. Knowing $f(0)=2$, let $a=0,b=0.1$. We have $\frac{1}{1.0001}\leq \frac{f(0.1)-2}{0.1-0} \leq1$, which gives $2+\frac{1}{10.001}\leq f(0.1) \leq2.1$ or $2.09999\leq f(0.1) \leq2.1$
