Question

does there exist a continuous function f(x) such that f(0)=10, f(2)=2 and f(x)≤1 for all x in (0,2)? answer ?

Explanation:

Step1: Apply the Mean - Value Theorem

By the Mean - Value Theorem, if \(y = f(x)\) is continuous on the closed interval \([a,b]=[0,2]\) and differentiable on the open interval \((a,b)=(0,2)\), then \(f(b)-f(a)=f'(c)(b - a)\) for some \(c\in(a,b)\). Here, \(a = 0\), \(b = 2\), \(f(0)=10\) and \(f(2)=2\). So \(f(2)-f(0)=f'(c)(2 - 0)\).

Step2: Calculate the value of \(f'(c)\)

Substitute the values of \(f(0)\) and \(f(2)\) into the Mean - Value Theorem formula: \(2-10=f'(c)\times2\). Simplify the left - hand side: \(- 8 = 2f'(c)\). Then solve for \(f'(c)\): \(f'(c)=-4\).

Step3: Analyze the result

We are given that \(f'(x)\leq1\) for all \(x\in(0,2)\). But from the Mean - Value Theorem, we found that there exists a \(c\in(0,2)\) such that \(f'(c)=-4\) which satisfies \(f'(c)\leq1\). Also, we can construct a linear function \(y=f(x)=10 - 4x\) which is continuous on \([0,2]\), \(f(0)=10\), \(f(2)=10-4\times2 = 2\) and \(f'(x)=-4\leq1\) for all \(x\in(0,2)\). So, yes, such a continuous function exists.

Answer:

Yes