Question

Let f be continuous on [a, b], differentiable on (a, b) and positive (i.e., > 0) for all x∈(a, b). Prove that there exists c∈(a, b) such that f′(c)/f(c)= (1/a-c)+(1/b−c). (Hint: consider the function F(x)=(x−a)(x−b)f(x) and use MVT for F(x) to show the existence of such c∈(a, b).)

Answer 1

To prove the given statement, we can use the Mean Value Theorem for a function F(x) and show the existence of c such that f'(c)/f(c) follows the desired form. By applying the MVT and simplifying the equation, we can ultimately express the desired equation.

Step-by-step explanation:

To prove that there exists a c in the interval (a, b) such that f'(c)/f(c) = (1/(a-c)) + (1/(b-c)), we can use the Mean Value Theorem (MVT) for the function F(x) = (x-a)(x-b)f(x).

1. Consider the function F(x) = (x-a)(x-b)f(x).
2. Since f is continuous on [a, b] and differentiable on (a, b), F(x) satisfies the conditions for applying the MVT.
3. According to the MVT, there exists a c in the interval (a, b) such that F'(c) = f'(c)(b-a+f(c)) = F(b) - F(a).
4. Plug in the values of F(b) = 0 and F(a) = (a-b)f(a) into the equation from step 3 to get f'(c)(b-a+f(c)) = (a-b)f(a).
5. Simplify the equation and divide both sides by f(c) to obtain f'(c)/f(c) = (a-b)/(b-a) = 1/((b-c)(a-c)).
6. Finally, use algebraic manipulation to express the right-hand side of the equation in the desired form (1/(a-c)) + (1/(b-c)).

Therefore, there exists a c in the interval (a, b) such that f'(c)/f(c) = (1/(a-c)) + (1/(b-c)).