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Question Number 103659 by abony1303 last updated on 16/Jul/20

When y=ax+b is a tangent line to the curve f(x)=x^3 passing through (0; −2), find a+b?

$$\mathrm{When}\:\mathrm{y}=\mathrm{ax}+\mathrm{b}\:\mathrm{is}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} \:\mathrm{passing}\:\mathrm{through}\:\left(\mathrm{0};\:−\mathrm{2}\right), \\ $$$$\mathrm{find}\:\mathrm{a}+\mathrm{b}? \\ $$

Commented by abony1303 last updated on 16/Jul/20

pls help

$$\mathrm{pls}\:\mathrm{help} \\ $$

Answered by Aziztisffola last updated on 16/Jul/20

y(0)=−2 ⇒a×0+b=−2⇒b=−2 Slope of (y=ax+b) is a and (y)∩(f)=(x_0 ,f(x_0 )) we get a=f ′(x_0 )=3x_0 ^2 { ((f(x_0 )=x_0 ^3 )),((y(x_0 )=ax_0 −2)) :} and f(x_0 )=y(x_0 )⇒ a=((x_0 ^3 +2)/x_0 )=3x_0 ^2 ⇒ 2x_0 ^3 −2=0 ⇒x_0 ^3 −1=0 ⇒x_0 =1 ⇒ a=3×1^2 =3 Hence y(x)=3x−2 ■

$$\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{2}\:\Rightarrow\mathrm{a}×\mathrm{0}+\mathrm{b}=−\mathrm{2}\Rightarrow\mathrm{b}=−\mathrm{2} \\ $$$$\:\mathrm{Slope}\:\mathrm{of}\:\left(\mathrm{y}=\mathrm{ax}+\mathrm{b}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{and} \\ $$$$\left(\mathrm{y}\right)\cap\left(\mathrm{f}\right)=\left(\mathrm{x}_{\mathrm{0}} ,\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)\right)\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}=\mathrm{f}\:'\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{3x}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\begin{cases}{\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} }\\{\mathrm{y}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{ax}_{\mathrm{0}} −\mathrm{2}}\end{cases}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}_{\mathrm{0}} \right)=\mathrm{y}\left(\mathrm{x}_{\mathrm{0}} \right)\Rightarrow \\ $$$$\mathrm{a}=\frac{\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} +\mathrm{2}}{\mathrm{x}_{\mathrm{0}} }=\mathrm{3x}_{\mathrm{0}} ^{\mathrm{2}} \:\Rightarrow\:\mathrm{2x}_{\mathrm{0}} ^{\mathrm{3}} −\mathrm{2}=\mathrm{0}\:\Rightarrow\mathrm{x}_{\mathrm{0}} ^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{0}} =\mathrm{1}\:\:\Rightarrow\:\mathrm{a}=\mathrm{3}×\mathrm{1}^{\mathrm{2}} =\mathrm{3} \\ $$$$\:\mathrm{Hence}\:\mathrm{y}\left(\mathrm{x}\right)=\mathrm{3x}−\mathrm{2}\:\blacksquare \\ $$

Answered by bemath last updated on 16/Jul/20

(1) (0,−2) ⇒−2=b ; y = ax−2 let (x_o , x_o ^3 ) the point in tangent line so ⇒slope = 3x_o ^2 = a and ((x_o ^3 +2)/x_o ) = a ⇔x_o ^3 +2 = 3x_o ^3 ⇔x_o =1 then the line y = ax−b passes through the point (1, 1) ⇒1 = a.1−2 , → { ((a=3)),((b=−2)) :} ∴ a+b = 1

$$\left(\mathrm{1}\right)\:\left(\mathrm{0},−\mathrm{2}\right)\:\Rightarrow−\mathrm{2}={b}\:;\:{y}\:=\:{ax}−\mathrm{2} \\ $$$${let}\:\left({x}_{{o}} ,\:{x}_{{o}} ^{\mathrm{3}} \right)\:{the}\:{point}\:{in}\:{tangent} \\ $$$${line}\:{so}\:\Rightarrow{slope}\:=\:\mathrm{3}{x}_{{o}} ^{\mathrm{2}} \:=\:{a}\:{and}\:\frac{{x}_{{o}} ^{\mathrm{3}} +\mathrm{2}}{{x}_{{o}} }\:=\:{a} \\ $$$$\Leftrightarrow{x}_{{o}} ^{\mathrm{3}} +\mathrm{2}\:=\:\mathrm{3}{x}_{{o}} ^{\mathrm{3}} \:\Leftrightarrow{x}_{{o}} =\mathrm{1}\:{then}\:{the}\:{line}\:{y}\:=\:{ax}−{b}\:{passes}\:{through}\: \\ $$$${the}\:{point}\:\left(\mathrm{1},\:\mathrm{1}\right)\:\Rightarrow\mathrm{1}\:=\:{a}.\mathrm{1}−\mathrm{2}\:,\:\rightarrow\begin{cases}{{a}=\mathrm{3}}\\{{b}=−\mathrm{2}}\end{cases} \\ $$$$\therefore\:{a}+{b}\:=\:\mathrm{1}\: \\ $$

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