iit kanpur Calculas practice question
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Does there exist a differentiable function f : [0, 2] → R satisfying f(0) = −1, f(2) = 4 and f’(x) ≤ 2 for all x ∈ [0, 2]?
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No. Suppose there is a function $f:[0,2] \rightarrow \mathbb{R}$ such that $f(0) = -1$ and $f(2) = 4$ and this $f$ is continuous on $[0,2]$ and differentiable in $(0,2)$ then according to mean value theorem there exists some $c$ in $(0,2)$ such that $f’ (c) = \frac{f(2)-f(0)}{2-0} = \frac{5}{2} > 2$ It means there is no $x$ such that $f’(x) \leq 2$ for the interval $[0,2]$
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There may exists x in the range [0,2]->$\Re$ for which f’(x)<=2 ,but for not all x as will be at least one x for which f’(x)=2.5 >2 . In the question they have said for all x in [0,2] , f’(x)<=2 that’s why the answer is no. Am i correct ? @ankitgupta.1729

 

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yes. it is correct. we can check graphically by joining two points (0,-1) and (2,4) and between then we can make graph such that slope of  tangent = $\tan \theta \in \mathbb{R}$ which can be $\leq 2$ for some points by making some portion of $f$ as increasing/decreasing for some interval.
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