Question

If $f\left(x\right)$ is continuous and differentiable over $[-2,5]$ and $-4\le f^{\prime }\left(x\right)\le 3$ for all $x$ in $(-2,5)$ then the greatest possible value of $f\left(5\right)-f(-2)$ is

• A. $7$
• B. $9$
• C. $15$
• D. $21$

Answer 1

Answer: (D)

$21$

f(x) is continuous on [-2,5] and differentiable on (-2,5).
So, by Lagrange's mean value theorem on [-2,5],
$f^{\prime }\left(x\right)=\frac{f\left(5\right)-f(-2)}{7}$
$\Rightarrow f\left(5\right)-f(-2)=7f^{\prime }\left(x\right)$

Since, $-4\le f^{\prime }\left(x\right)\le 3$
$-28\le 7f^{\prime }\left(x\right)\le 21$
$-28\le f\left(5\right)-f(-2)\le 21$