What is the equation of the normal line of $f (x) = {(x - 2)}^{\frac{3}{2}} - x^{3}$ $f(x)=(x-2)^(3/2)-x^3$ at $x = 2$ $x=2$ ?

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Answer 1 · 2016-01-10T11:07:49

The Normal Line is

$y = \frac{x}{12} - \frac{49}{6}$ $y=x/12-49/6$ or $x - 12 y = 98$ $x-12y=98$

the given: $f (x) = {(x - 2)}^{\frac{3}{2}} - x^{3}$ $f(x) =(x-2)^(3/2)-x^3$ and $x = 2$ $x=2$

$f (2) = {(2 - 2)}^{\frac{3}{2}} - {(2)}^{3}$ $f(2) =(2-2)^(3/2)-(2)^3$

$f (2) = - 8$ $f(2)= -8$

the point on the curve : $(2, - 8)$ $(2, -8)$

compute slope $m$ $m$ then use $- \frac{1}{m}$ $-1/m$ for the normal line

$f' (x) = (3/2)(x-2)^(1/2)-3x^2$

$f' (2) = (3/2)(2-2)^(1/2)-3(2)^2$

$f' (2) = -12$

The slope to be used to find the normal line is $-1/m$

$-1/m=-1/-12=1/12$

Determine now the Normal Line using

$y-y_1=(1/12)(x-x_1)$

$y-(-8)=(1/12)(x-2)$

after simplification the final answer:

$x-12y=98$ or $y=x/12-49/6$

What is the equation of the normal line of f(x)=(x-2)^(3/2)-x^3f(x)=(x−2)32−x3f(x)=(x-2)^(3/2)-x^3 at x=2x=2x=2?