1 votes 1 votes 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? Nirmal Gaur asked Sep 22, 2019 Nirmal Gaur 1.4k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply `JEET commented Sep 22, 2019 reply Follow Share can possibly have? What? Is the question complete? 0 votes 0 votes Nirmal Gaur commented Sep 22, 2019 reply Follow Share Yes the question is complete. You can check it here: https://brilliant.org/wiki/mean-value-theorem/ 0 votes 0 votes Nirmal Gaur commented Sep 22, 2019 reply Follow Share 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 votes 1 votes techbd123 commented Oct 8, 2019 reply Follow Share @Nirmal Gaur You can post it as the answer instead of a comment. 0 votes 0 votes Nirmal Gaur commented Oct 9, 2019 reply Follow Share @ techbd123 Done :) 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes 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. Nirmal Gaur answered Oct 9, 2019 • edited Oct 10, 2019 by Lakshman Bhaiya Nirmal Gaur comment Share Follow See all 0 reply Please log in or register to add a comment.