When-y-ax-b-is-a-tangent-line-to-the-curve-f-x-x-3-passing-through-0-2-find-a-b-

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 by abony1303 last updated on 16/Jul/20

pls help

Answered by Aziztisffola last updated on 16/Jul/20

y (0) = - 2 \Rightarrow a \times 0 + b = - 2 \Rightarrow b = - 2

Slope of (y = ax + b) is a and

(y) \cap (f) = (x_{0}, f (x_{0})) we get a = f^{'} (x_{0}) = {3 x}_{0}^{2}

{\begin{cases} f (x_{0}) = x_{0}^{3} \\ y (x_{0}) = {ax}_{0} - 2 \end{cases} and f (x_{0}) = y (x_{0}) \Rightarrow

a = \frac{x_{0}^{3} + 2}{x_{0}} = {3 x}_{0}^{2} \Rightarrow {2 x}_{0}^{3} - 2 = 0 \Rightarrow x_{0}^{3} - 1 = 0

\Rightarrow x_{0} = 1 \Rightarrow a = 3 \times 1^{2} = 3

Hence y (x) = 3 x - 2

Answered by bemath last updated on 16/Jul/20

(1) (0, - 2) \Rightarrow - 2 = b; y = a x - 2

l e t (x_{o}, x_{o}^{3}) t h e p o i n t i n t a n g e n t

l i n e s o \Rightarrow s l o p e = 3 x_{o}^{2} = a a n d \frac{x_{o}^{3} + 2}{x_{o}} = a

\Leftrightarrow x_{o}^{3} + 2 = 3 x_{o}^{3} \Leftrightarrow x_{o} = 1 t h e n t h e l i n e y = a x - b p a s s e s t h r o u g h

t h e p o i n t (1, 1) \Rightarrow 1 = a . 1 - 2, \to {\begin{cases} a = 3 \\ b = - 2 \end{cases}

∴ a + b = 1

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