Answer
$1.09999\leq f(0.1) \leq 1.1$
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. Graph the given function $f'(x)=\frac{1}{1+x^4cos(x)}, 0\leq x\leq 0.1$
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 $min\ f'=f'(0.1)=\frac{1}{1+0.1^4cos(0.1)}\approx0.9999$ and $max\ f'=f'(0)=\frac{1}{1+0^4cos(0)}=1$
Step 4. Knowing $f(0)=1$, let $a=0, b=0.1$. We have $0.9999\leq \frac{f(0.1)-f(0)}{0.1-0} \leq 1$ which gives $1.09999\leq f(0.1) \leq 1.1$
