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If $f(1)=10$ and $f '(x)\geq 2 $ for $1\leq x \leq 4 $ how small can $f(4)$ possibly be?

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Toolbox:
  • Lagrange's Mean Value Theorem :
  • Let $f(x) $ be a real valued function that satisfies the following conditions.
  • (i) $f(x)$ is continuous on the closed interval $[a,b]$
  • (ii) $f(x)$ is differentiable in the open interval $(a,b)$
  • (iii) $f(a)=f(b)$
  • Then there exists atleast one value $c \in (a,b)$ such that $f'(c)=0$
$f(1)=10, f'(x) \geq 2, 1 \leq x \leq 4$
Step 1:
Assuming the condition's for Lagrange's theorem are satisfied in $[1,4]$ there is a $ c \in (1,4)$ Such that $f'(c)=\large\frac{f(4)-f(1)}{4-1}$
Step 2:
Now $f'(c) \geq \qquad \therefore \large\frac{f(4)-10}{3} $$ \geq 2$
$f(4) \geq 16$
The smallest values that $f(4)$ can take is $16$
answered Jul 25, 2013 by meena.p
 

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