Let f (x) be differentiable function in [0, 3]. If f (0) = 0 and \left|\right. \textrm{ } f ' \left(\right. x \left.\right) \textrm{ } \left|\right. \textrm{ }\textrm{ } \leq \textrm{ }\textrm{ } \frac{1}{3} then

Engineering

Mathematics

Lagranges Mean Value Theorem

Question

Let f (x) be differentiable function in [0, 3]. If f (0) = 0 and $| f' (x) | \leq \frac{1}{3}$ then

f (x) = 3x

f (x) $\leq$ 3

f (x) = 4 for at least one x in x $\in$ [0, 2]

|f(x)| $\leq$ 1

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$| f' (x) | = | \frac{f (x) - f (0)}{x - 0} | \leq \frac{1}{3}$

$\Rightarrow | f (x) | \leq \frac{1}{3} | x | \leq \frac{1}{3} \times 3 \Rightarrow | f' (x) | \leq 1$