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Exercise 7.3 · Q8

Q.Does there exist a differentiable function f(x)f(x) such that f(0)=−1,f(2)=4f(0)=-1,\\ f(2)=4 and f′(x)le2f'(x)\\le2 for all xx. Justify your answer.

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Assume such an ff exists and derive a contradiction from LMVT.

Step 1. Suppose such a differentiable ff exists.

f(0)=−1f(0)=-1, f(2)=4f(2)=4, and f′(x)≤2f'(x)\le2 for all xx.

Step 2. Apply LMVT on [0,2][0,2].

Since ff is differentiable everywhere (in particular on [0,2][0,2]), LMVT guarantees some c∈(0,2)c\in(0,2) with

f′(c)=f(2)−f(0)2−0=4−(−1)2=52=2.5.f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{4-(-1)}{2}=\frac{5}{2}=2.5.

Step 3. Compare with the given bound.

This forces f′(c)=2.5f'(c)=2.5, but the hypothesis states f′(x)≤2f'(x)\le2 for all xx — in particular f′(c)≤2f'(c)\le2. Since 2.5>22.5>2, this is a contradiction. …

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