First slide

Mean value theorems

Question

$If a < (28)^{\frac{1}{3}} - 3 < b, then (a, b) is$

Moderate

A

$(\frac{1}{28}, \frac{1}{27})$
B

$(\frac{1}{28}, \frac{1}{29})$
C

$(27, 28)$
D

$(\frac{1}{27}, \frac{1}{26})$

Solution

$Let f (x) = x^{\frac{1}{3}}, x \in [27, 28]$
Differentiating with respect to $x$ on both sides
$\Rightarrow f' (x) = \frac{1}{3 x^{\frac{2}{3}}}$
$By Lagrange's Mean value theorem: \frac{f (b) - f (a)}{b - a} = f^{'} (c), where c \in (27, 28)$
$\begin{array}{l} \Rightarrow \frac{(28)^{\frac{1}{3}} - (27)^{\frac{1}{3}}}{28 - 27} = f^{'} (c) \\ \Rightarrow (28)^{\frac{2}{3}} - 3 = \frac{1}{3 c^{\frac{2}{3}}} \\ Now, 27 < c < 28 \\ \Rightarrow 9 < c^{\frac{2}{3}} < (28)^{\frac{2}{3}} \end{array}$
$\begin{array}{l} \Rightarrow \frac{1}{9} > \frac{1}{c^{\frac{2}{3}}} > \frac{1}{28^{\frac{2}{3}}} \\ \Rightarrow \frac{1}{27} > \frac{1}{3 . c^{\frac{2}{3}}} > \frac{1}{3 (28)^{\frac{2}{3}}} = \frac{(28)^{\frac{1}{3}}}{3} \frac{1}{28} > \frac{1}{28} \\ ∴ \frac{1}{28} < \frac{1}{3 (c)^{\frac{2}{3}}} < \frac{1}{27} \\ \Rightarrow \frac{1}{28} < (28)^{\frac{1}{3}} - 3 < \frac{1}{27} \end{array}$

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