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

Given ∣f(x)∣ = x^2 for −1<x<1 find f ′(0).

$$\:\mathrm{Given}\:\mid\mathrm{f}\left(\mathrm{x}\right)\mid\:=\:\mathrm{x}^{\mathrm{2}} \:\mathrm{for}\:−\mathrm{1}<\mathrm{x}<\mathrm{1} \\ $$ $$\mathrm{find}\:\mathrm{f}\:'\left(\mathrm{0}\right). \\ $$

Commented byEDWIN88 last updated on 28/Feb/21

does not exist

$$\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\ $$

Commented byEDWIN88 last updated on 28/Feb/21

Commented byMJS_new last updated on 28/Feb/21

proper maths: let′s try with f(x)=2x ∀ x∈R this simply means, we′re assigning x → 2x or y=2x ∣f(x)∣=2x is not a function anymore ∣y∣=2x ⇔ y=±2x∧x≥0 we can split the branches: y_1 =−2x∧x≥0 y_2 =2x∧x≥0 (dy_1 /dx) and (dy_2 /dx) both exist but (dy/dx) doesn′t exist now ∣f(x)∣=x^2 ∀ x∈R y=±x^2 ∧x^2 ≥0 but x^2 ≥0∀x∈R again we have 2 branches: y_1 =−x^2 ∀ x∈R y_2 =x^2 ∀ x∈R again (dy_1 /dx) and (dy_2 /dx) exist but (dy/dx) doesn′t exist anyway for x=0 we have (dy_1 /dx)=(dy_2 /dx)=0 but this still doesn′t mean that (dy/dx)=0 for x=0

$$\mathrm{proper}\:\mathrm{maths}: \\ $$ $$\mathrm{let}'\mathrm{s}\:\mathrm{try}\:\mathrm{with} \\ $$ $${f}\left({x}\right)=\mathrm{2}{x}\:\forall\:{x}\in\mathbb{R} \\ $$ $$\:\:\:\:\:\mathrm{this}\:\mathrm{simply}\:\:\mathrm{means},\:\mathrm{we}'\mathrm{re}\:\mathrm{assigning}\:{x}\:\rightarrow\:\mathrm{2}{x} \\ $$ $$\:\:\:\:\:\mathrm{or}\:{y}=\mathrm{2}{x} \\ $$ $$\mid{f}\left({x}\right)\mid=\mathrm{2}{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{function}\:\mathrm{anymore} \\ $$ $$\:\:\:\:\:\mid{y}\mid=\mathrm{2}{x}\:\Leftrightarrow\:{y}=\pm\mathrm{2}{x}\wedge{x}\geqslant\mathrm{0} \\ $$ $$\:\:\:\:\:\mathrm{we}\:\mathrm{can}\:\mathrm{split}\:\mathrm{the}\:\mathrm{branches}: \\ $$ $$\:\:\:\:\:{y}_{\mathrm{1}} =−\mathrm{2}{x}\wedge{x}\geqslant\mathrm{0} \\ $$ $$\:\:\:\:\:{y}_{\mathrm{2}} =\mathrm{2}{x}\wedge{x}\geqslant\mathrm{0} \\ $$ $$\:\:\:\:\:\frac{{dy}_{\mathrm{1}} }{{dx}}\:\mathrm{and}\:\frac{{dy}_{\mathrm{2}} }{{dx}}\:\mathrm{both}\:\mathrm{exist}\:\mathrm{but}\:\frac{{dy}}{{dx}}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$ $$ \\ $$ $$\mathrm{now} \\ $$ $$\mid{f}\left({x}\right)\mid={x}^{\mathrm{2}} \:\forall\:{x}\in\mathbb{R} \\ $$ $$\:\:\:\:\:{y}=\pm{x}^{\mathrm{2}} \wedge{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\mathrm{but}\:{x}^{\mathrm{2}} \geqslant\mathrm{0}\forall{x}\in\mathbb{R} \\ $$ $$\:\:\:\:\:\mathrm{again}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{branches}: \\ $$ $$\:\:\:\:\:{y}_{\mathrm{1}} =−{x}^{\mathrm{2}} \:\forall\:{x}\in\mathbb{R} \\ $$ $$\:\:\:\:\:{y}_{\mathrm{2}} ={x}^{\mathrm{2}} \:\forall\:{x}\in\mathbb{R} \\ $$ $$\:\:\:\:\:\mathrm{again}\:\frac{{dy}_{\mathrm{1}} }{{dx}}\:\mathrm{and}\:\frac{{dy}_{\mathrm{2}} }{{dx}}\:\mathrm{exist}\:\mathrm{but}\:\frac{{dy}}{{dx}}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$ $$\:\:\:\:\:\mathrm{anyway}\:\mathrm{for}\:{x}=\mathrm{0}\:\mathrm{we}\:\mathrm{have}\:\frac{{dy}_{\mathrm{1}} }{{dx}}=\frac{{dy}_{\mathrm{2}} }{{dx}}=\mathrm{0} \\ $$ $$\:\:\:\:\:\mathrm{but}\:\mathrm{this}\:\mathrm{still}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:\mathrm{that}\:\frac{{dy}}{{dx}}=\mathrm{0}\:\mathrm{for}\:{x}=\mathrm{0} \\ $$

Answered by EDWIN88 last updated on 28/Feb/21

∣y ∣ = x^2 ⇒ (√y^2 ) =x^2 ; y^2 = x^4 2y y′ = 4x^3 , y′ = ((2x^3 )/( (√y))) = (0/0) (does not exist )

$$\mid\mathrm{y}\:\:\mid\:=\:\mathrm{x}^{\mathrm{2}} \:\Rightarrow\:\sqrt{\mathrm{y}^{\mathrm{2}} }\:=\mathrm{x}^{\mathrm{2}} \:;\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}^{\mathrm{4}} \\ $$ $$\mathrm{2y}\:\mathrm{y}'\:=\:\mathrm{4x}^{\mathrm{3}} \:,\:\mathrm{y}'\:=\:\frac{\mathrm{2x}^{\mathrm{3}} }{\:\sqrt{\mathrm{y}}}\:=\:\frac{\mathrm{0}}{\mathrm{0}}\:\left(\mathrm{does}\:\mathrm{not}\:\mathrm{exist}\:\right) \\ $$

Commented bymr W last updated on 28/Feb/21

y′ = ((2x^3 )/( (√y))) =((2x^3 )/x^2 )=2x f′(0)=0

$$\mathrm{y}'\:=\:\frac{\mathrm{2x}^{\mathrm{3}} }{\:\sqrt{\mathrm{y}}}\:=\frac{\mathrm{2}{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} }=\mathrm{2}{x} \\ $$ $${f}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$

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