Let f(x) be a function satisfying |f(x)|≤x2 for −1≤x≤1, how do you show that f is differentiable at x = 0 and find f’(0)?

Since |f(x)|≤x2 for ξn[−1,1], we must have|f(0)|≤02=0, but since it is definitely non-negative, it must be 0.f(0)=0Nowf′(0)=limh→0f(0+h)−f(0)h=limh→0f(h)hThus|f′(0)|=limh→0|f(h)h|=limh→0f(h)h≤limh→0h2h=limh→0h=0Hence |f′(0)|≤0⇒|f′(0)|=0Hence the derivative at x=0 exists and is 0.

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High School
Calculus and Analysis
Calculus 1
Derivatives

Zane Decker

Answered question

2022-04-15

Let f(x) be a function satisfying

| f (x) | \leq x^{2}

for

- 1 \leq x \leq 1

, how do you show that f is differentiable at x = 0 and find f’(0)?

Answer & Explanation

zakos2zn1mr

Beginner2022-04-16Added 10 answers

Since

| f (x) | \leq x^{2}

for

ξ n [- 1, 1]

, we must have

| f (0) | \leq 0^{2} = 0

, but since it is definitely non-negative, it must be 0.
f(0)=0
Now

f^{'} (0) = lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = lim_{h \to 0} \frac{f (h)}{h}

Thus

| f^{'} (0) | = lim_{h \to 0} | \frac{f (h)}{h} | = lim_{h \to 0} \frac{f (h)}{h}

\leq lim_{h \to 0} \frac{h^{2}}{h} = lim_{h \to 0} h = 0

Hence

| f^{'} (0) | \leq 0 \Rightarrow | f^{'} (0) | = 0

Hence the derivative at x=0 exists and is 0.

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