can possibly have?

What?

Is the question complete?

What?

Is the question complete?

1 vote

f(x) is a differentiable function that satisfies 5 ≤ f′(x) ≤ 14 for all x. Let a and b be the maximum and minimum values, respectively, that f(11)−f(3) can possibly have, then what is the value of a+b?

1

Solution:

Since **f(x)** is differentiable on all intervals, we can choose any two points (Here we take 11 and 3). So from the mean value theorem, we have some c such that

$f'(c) = (f(11)-f(3))/11-3$)

Now given $5 \leqslant f'(c)\leqslant14$ therefore $5 \leqslant (f(11)-f(3))/8 \leqslant14$

=> $40 \leqslant (f(11)-f(3)) \leqslant112$

Hence a = 112 and b = 40 which gives a+b = **152** as the answer.

1 vote

Since $f(x)$ is differentiable on all intervals, we can choose any two points $($ Here we take $11$ and $3)$. So from the mean value theorem, we have some $c$ such that:

$f'\left ( c \right )=\dfrac{\bigg(f(11)-f(3)\bigg)}{(11-3)}$

Now given $5\leqslant f'\left ( c \right )\leqslant 14$ therefore:

$\implies 5\leqslant \dfrac{\bigg(f(11)-f(3)\bigg)}{8}\leqslant 14$

$\implies 40\leqslant \bigg(f(11)-f(3)\bigg)\leqslant 112$

Hence $a = 112$ and $b = 40$ which gives $a+b = 152$ as the answer.

$f'\left ( c \right )=\dfrac{\bigg(f(11)-f(3)\bigg)}{(11-3)}$

Now given $5\leqslant f'\left ( c \right )\leqslant 14$ therefore:

$\implies 5\leqslant \dfrac{\bigg(f(11)-f(3)\bigg)}{8}\leqslant 14$

$\implies 40\leqslant \bigg(f(11)-f(3)\bigg)\leqslant 112$

Hence $a = 112$ and $b = 40$ which gives $a+b = 152$ as the answer.