Given-f-x-x-2-for-1-lt-x-lt-1-find-f-0-

Question Number 134127 by bobhans last updated on 28/Feb/21

Given ∣ f (x) ∣ = x^{2} for - 1 < x < 1

find f^{'} (0) .

Commented by EDWIN88 last updated on 28/Feb/21

does not exist

Commented by EDWIN88 last updated on 28/Feb/21

Commented by MJS_new last updated on 28/Feb/21

proper maths :

{let}^{'} s try with

f (x) = 2 x \forall x \in R

this simply means, {we}^{'} re assigning x \to 2 x

or y = 2 x

∣ f (x) ∣= 2 x is not a function anymore

∣ y ∣= 2 x \Leftrightarrow y = \pm 2 x \land x ⩾ 0

we can split the branches :

y_{1} = - 2 x \land x ⩾ 0

y_{2} = 2 x \land x ⩾ 0

\frac{{d y}_{1}}{d x} and \frac{{d y}_{2}}{d x} both exist but \frac{d y}{d x} {doesn}^{'} t exist

now

∣ f (x) ∣= x^{2} \forall x \in R

y = \pm x^{2} \land x^{2} ⩾ 0 but x^{2} ⩾ 0 \forall x \in R

again we have 2 branches :

y_{1} = - x^{2} \forall x \in R

y_{2} = x^{2} \forall x \in R

again \frac{{d y}_{1}}{d x} and \frac{{d y}_{2}}{d x} exist but \frac{d y}{d x} {doesn}^{'} t exist

anyway for x = 0 we have \frac{{d y}_{1}}{d x} = \frac{{d y}_{2}}{d x} = 0

but this still {doesn}^{'} t mean that \frac{d y}{d x} = 0 for x = 0

Answered by EDWIN88 last updated on 28/Feb/21

∣ y ∣ = x^{2} \Rightarrow \sqrt{y^{2}} = x^{2}; y^{2} = x^{4}

2 y y^{'} = {4 x}^{3}, y^{'} = \frac{{2 x}^{3}}{\sqrt{y}} = \frac{0}{0} (does not exist)

Commented by mr W last updated on 28/Feb/21

y^{'} = \frac{{2 x}^{3}}{\sqrt{y}} = \frac{2 x^{3}}{x^{2}} = 2 x

f^{'} (0) = 0

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