Not functions again

Algebra Level 4

Let |f(x)-f(y)|<= ∣ x − y ∣ 3 |x-y|^{3} for all x,y belongs to R

Then the value of f'(x) is

Details and Assumptions

|.| denotes absolute value or mod function and R is the set of real no.

1 0 Not possible ∞ \infty -1 Can't be determined

Naman Kapoor by Naman Kapoor , 22, India

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1 solution

Replace x → x + h x \to x + h and y → x y \to x

⇒ ∣ f ( x + h ) − f ( x ) ∣ ≤ ( x + h − x ) 3 ⇒ ∣ f ( x + h ) − f ( x ) ∣ ≤ h 3 \Rightarrow |f(x+h) - f(x)| \leq (x + h - x)^{3} \Rightarrow |f(x+h) -f(x)| \leq h^{3}

⇒ − h 3 ≤ f ( x + h ) − f ( x ) ≤ h 3 \Rightarrow -h^{3} \leq f(x+h) - f(x) \leq h^{3}

⇒ − h 2 ≤ f ( x + h ) − f ( x ) h ≤ h 2 \Rightarrow -h^{2} \leq \dfrac{f(x+h) - f(x)}{h} \leq h^{2}

⇒ lim ⁡ h → 0 − h 2 ≤ lim ⁡ h → 0 f ( x + h ) − f ( x ) h ≤ lim ⁡ h → 0 h 2 \Rightarrow \displaystyle \lim_{h\to 0} -h^{2} \leq \lim_{h \to 0} \dfrac{f(x+h) -f(x)}{h} \leq \lim_{h\to 0} h^{2}

⇒ 0 ≤ f ′ ( x ) ≤ 0 ⇒ f ′ ( x ) = 0 \Rightarrow 0 \leq f'(x) \leq 0 \Rightarrow f'(x) = 0

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