My notation may not be 'grammatically' correct, and my process is probably iffy. I'm only in calc 1 and would appreciate if someone could check my work and possibly correct me on the right process's and words to use.
The problem states:
- $f(x)$ continuous on $[0,5]$ (1)
- $f(0) = 1$ (2)
- $f'(x) \in [4,8] \forall x \in (0,5)$ (3)
Then asks you to determine if $f(5) \in [20,30]$.
Work
- Assume $f(5) \in [20,30]$.
- $\frac{20 - f(0)}{5 - 0}\leq \frac{f(5)-f(0)}{5 - 0}\leq \frac{30 - f(0)}{5 - 0}$, expanding and transforming the interval into an inequality
- $\frac{19}{5} \leq \frac{f(5)-f(0)}{5 - 0}\leq \frac{29}{5}$, simplifying inequality
- $\frac{19}{5} \leq f'(c) \leq \frac{29}{5} | c \in (0,5)$, since MVT guarantees $\frac{f(b)-f(a)}{b - a} = f'(c)$ for a value $c \in (a,b)$.
- Let $f'(c)$ equal $\frac{19.5}{5} = 3.9$ which satisfies the inequality derived in 4. But it violates the 3rd given, $f'(x) \in [4,8] \forall x \in (0,5)$. Therefore, $f(5) \notin [20,30]$.
Thanks.