Question

Let the function, $f:[-7,0]\to R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and ^′$f\text{'}\left(x\right)\le 2$, for all $\in (-7,0)$, then for all such functions, $f(-1)+f\left(0\right)$ lies in the interval:

• A. $[-6,20]$
• B. $(-\infty ,20]$
• C. $(-\infty ,11]$
• D. $[-3,11]$

Answer 1

The correct option is B

$(-\infty ,20]$

Determine the interval for the function:

We have, $f(-7)=-3andf\text{'}\left(x\right)\le 2$

Applying Lagrange's mean value theorem in $[-7,0]$, we get

$$\begin{gathered} \frac{\left(f\right(-7)–f(0\left)\right)}{-7}=f\text{'}\left(c\right)\le 2 \\ \Rightarrow \frac{(-3-f(0\left)\right)}{-7}\le 2 \\ \Rightarrow f\left(0\right)+3\le 14 \\ \Rightarrow f\left(0\right)\le 11 \end{gathered}$$ $\left[\because f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\right]$

Applying Lagrange's mean value theorem in $[-7,-1]$, we get

$$\begin{gathered} \frac{\left(f\right(-7)–f(-1\left)\right)}{(-7+1)}=f\text{'}\left(c\right)\le 2 \\ \frac{-3–f(-1))}{-6}=f\text{'}\left(c\right)\le 2 \\ f(-1)+3=\le 12 \\ f(-1)\le 9 \end{gathered}$$

Hence,$f(-1)+f\left(0\right)\le 20$

Therefore, option (B) is the correct answer.