Question 38

Let f be a differentiable function on [-2, 2] such that f(-2) = 1, f(2) = 5 and $$\mid \frac{df(x)}{dx}\mid \leq 1$$ for all $$x \epsilon [-2, 2]$$. The value of f(0) is

Solution

Given that $$\left|\ \frac{\ df\left(x\right)}{dx}\right|\le\ 1$$

Hence $$-1\le\ \ \frac{\ df\left(x\right)}{dx}\le\ 1$$. Taking definite intergral from [-2 ,2]

-1*(5 - 1) $$\le\ $$ f(2) - f(-2) $$\le\ $$ (5-1)

$$-4\le\ 4\le\ 4$$. It is only possible when $$\ \frac{\ df\left(x\right)}{dx}$$ is always equal to 1. Therefore f(x) is a straight line with slope 1

f(x) = x+3, f(0) = 3

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