Question:
Let the function, $f:[-7,0] \rightarrow R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f^{\prime}(x) \leq 2$, for all $x \in(-7,0)$, then for all such functions $f$, $f^{\prime}(-1)+f(0)$ lies in the interval:
Correct Option: 1
Solution:
From, LMVT for $x \in[-7,-1]$
$\frac{f(-1)-f(-7)}{(-1+7)} \leq 2 \Rightarrow \frac{f(-1)+3}{6} \leq 2 \Rightarrow f(-1) \leq 9$
From, LMVT for $x \in[-7,0]$
$\frac{f(0)-f(-7)}{(0+7)} \leq 2$
$\frac{f(0)+3}{7} \leq 2 \Rightarrow f(0) \leq 11$
$\therefore \quad f(0)+f(-1) \leq 20$