First slide

Mean value theorems

Question

$Suppose f is differentiable on R and a \leq f^{'} (x) \leq b for all x \in R where$ $a, b > 0 . If f (0) = 0, then$

Moderate

A

$f (x) \leq \min (a x, b x)$
B

$f (x) \geq \max (a x, b x)$
C

$a \leq f (x) \leq b$
D

$a x \leq f (x) \leq b x$

Solution

For x > 0. Applying Lagrange’s theorem on [0, x] we have $c \in (0, x)$
such that $\frac{f (x)}{x} = \frac{f (x) - f (0)}{x - 0} = f' (c)$
$\begin{array}{l} But a \leq f^{'} (c) \leq b so a \leq \frac{f (x)}{x} \leq b \Rightarrow a x \leq f (x) \leq b x, x > 0 \\ Similarly for x < \underline{0} applying Lagrange's theorem for [x, 0 ⌉, \\ we have a x \leq f (x) \leq b x \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App