Question-74235

Question Number 74235 by aliesam last updated on 20/Nov/19

Answered by Joel578 last updated on 20/Nov/19

∣x∣ = (√x^2 ) f(x) = (√((x^2 − 3)^2 )) f ′(x) = (1/(2(√((x^2 − 3)^2 )))) . 2(x^2 − 3) . 2x = ((2x(x^2 − 3))/( (√((x^2 − 3)^2 )))) = ((2x(x^2 − 3))/(∣x^2 − 3∣))

$$\mid{x}\mid\:=\:\sqrt{{x}^{\mathrm{2}} } \\ $$$${f}\left({x}\right)\:=\:\sqrt{\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${f}\:'\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)^{\mathrm{2}} }}\:.\:\mathrm{2}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)\:.\:\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)}{\:\sqrt{\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}{x}\left({x}^{\mathrm{2}} \:−\:\mathrm{3}\right)}{\mid{x}^{\mathrm{2}} \:−\:\mathrm{3}\mid} \\ $$

Commented by MJS last updated on 20/Nov/19

(a/(∣a∣))=((∣a∣)/a)=sign a usually sign 0 :=1 ⇒ we can write f′(x)=2x sign (x^2 −3)

$$\frac{{a}}{\mid{a}\mid}=\frac{\mid{a}\mid}{{a}}=\mathrm{sign}\:{a} \\ $$$$\mathrm{usually}\:\mathrm{sign}\:\mathrm{0}\::=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{write}\:{f}'\left({x}\right)=\mathrm{2}{x}\:\mathrm{sign}\:\left({x}^{\mathrm{2}} −\mathrm{3}\right) \\ $$

Commented by Joel578 last updated on 21/Nov/19

$${thank}\:{you}\:{Sir} \\ $$

Answered by MJS last updated on 20/Nov/19

∣x^2 −3∣=x^2 −3∧x^2 ≥3 ∨ ∣x^2 −3∣=3−x^2 ∧x^2 <3 f(x)= { ((3−x^2 ∧−(√3)≤x≤(√3))),((x^2 −3∧(x<−(√3)∨x>(√3)))) :} f′(x)= { ((−2x∧−(√3)≤x≤(√3))),((2x∧(x<−(√3)∨x>(√3)))) :}

$$\mid{x}^{\mathrm{2}} −\mathrm{3}\mid={x}^{\mathrm{2}} −\mathrm{3}\wedge{x}^{\mathrm{2}} \geqslant\mathrm{3}\:\vee\:\mid{x}^{\mathrm{2}} −\mathrm{3}\mid=\mathrm{3}−{x}^{\mathrm{2}} \wedge{x}^{\mathrm{2}} <\mathrm{3} \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{3}−{x}^{\mathrm{2}} \wedge−\sqrt{\mathrm{3}}\leqslant{x}\leqslant\sqrt{\mathrm{3}}}\\{{x}^{\mathrm{2}} −\mathrm{3}\wedge\left({x}<−\sqrt{\mathrm{3}}\vee{x}>\sqrt{\mathrm{3}}\right)}\end{cases} \\ $$$${f}'\left({x}\right)=\begin{cases}{−\mathrm{2}{x}\wedge−\sqrt{\mathrm{3}}\leqslant{x}\leqslant\sqrt{\mathrm{3}}}\\{\mathrm{2}{x}\wedge\left({x}<−\sqrt{\mathrm{3}}\vee{x}>\sqrt{\mathrm{3}}\right)}\end{cases} \\ $$

Commented by aliesam last updated on 20/Nov/19

$${thank}\:{you}\:{sir} \\ $$

Answered by arkanmath7@gmail.com last updated on 21/Nov/19

f(x) = ∣g(x)∣ f ′(x) = ((g(x))/(∣g(x)∣)) ∙ g^′ (x) let g(x) = x^2 −3 ⇒ g^′ (x) = 2x f^′ (x) = ((x^2 − 3)/(∣x^2 − 3∣)) ∙ 2x = ((2x^3 −6x)/(∣x^2 − 3∣))

$${f}\left({x}\right)\:=\:\mid{g}\left({x}\right)\mid \\ $$$${f}\:'\left({x}\right)\:=\:\frac{{g}\left({x}\right)}{\mid{g}\left({x}\right)\mid}\:\centerdot\:{g}\:^{'} \left({x}\right) \\ $$$${let}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:−\mathrm{3}\:\:\Rightarrow\:{g}\:^{'} \left({x}\right)\:=\:\mathrm{2}{x} \\ $$$${f}\:^{'} \left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} \:−\:\mathrm{3}}{\mid{x}^{\mathrm{2}} \:−\:\mathrm{3}\mid}\:\centerdot\:\mathrm{2}{x}\:=\:\frac{\mathrm{2}{x}^{\mathrm{3}} \:−\mathrm{6}{x}}{\mid{x}^{\mathrm{2}} \:−\:\mathrm{3}\mid} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply