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?

Commented byabony1303 last updated on 16/Jul/20

pls 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 ■

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