Let S be the set of all functions f: [0, 1]arrow R, which are continuous on [0, 1] and differentiable on (0,1). Then for every f in S, there exists a c ϵ (0, 1). depending on f, such that:

Question

Let S be the set of all functions f: [0, 1]arrow R, which are continuous on [0, 1] and differentiable on (0,1). Then for every f in S, there exists a c ϵ (0, 1). depending on f, such that: (A) (f (1)-f (c)/1-c)=f '(c) (B) |f (c)-f (1)| < |f '(c)| (C) |f (c)+f (1)| <(1+c) |f '(c)| (D) |f (c)-f (1)| <(1-c) |f '(c)|

Tardigrade Admin · Accepted Answer

Correct answer is (d) |f (c)-f (1)| <(1-c) |f '(c)|

Q. Let $S$ be the set of all functions $f : [0, 1] \to R,$ which are continuous on [0, 1] and differentiable on (0,1). Then for every $f$ in $S$ , there exists $a c ϵ (0, 1) .$ depending on $f$ , such that:

$\frac{f ( 1 ) - f ( c )}{1 - c} = f^{'} (c)$

$∣ f (c) - f (1) ∣ < ∣ f^{'} (c) ∣$

$∣ f (c) + f (1) ∣ < (1 + c) ∣ f^{'} (c) ∣$

$∣ f (c) - f (1) ∣ < (1 - c) ∣ f^{'} (c) ∣$

Correct answer is (d) $∣ f (c) - f (1) ∣ < (1 - c) ∣ f^{'} (c) ∣$

Q. Let S be the set of all functions f:[0,1]→R, which are continuous on [0, 1] and differentiable on (0,1). Then for every f in S, there exists acϵ(0,1). depending on f, such that:

1−cf(1)−f(c)​=f′(c)

∣f(c)−f(1)∣<∣f′(c)∣

∣f(c)+f(1)∣<(1+c)∣f′(c)∣

∣f(c)−f(1)∣<(1−c)∣f′(c)∣