What is the equation of the normal line of f(x)=1/x^2-xf(x)=1x2−x at x = 4x=4?

1 Answer
Vinícius Ferraz
Nov 15, 2015

y = 32/33 x - 4127/528 Leftrightarrow 528y = 512x - 4127y=3233x−4127528⇔528y=512x−4127

Explanation:

f(4) = 1/16 - 4 Rightarrow P = (4, -63/16)f(4)=116−4⇒P=(4,−6316)

(d)/(dx)f(x) = -2x^-3 - 1ddxf(x)=−2x−3−1

(d)/(dx)f(4) = -2/4^3 - 1 = (-1-32)/32 = mddxf(4)=−243−1=−1−3232=m // inclination of the tangent line

a = -1/m = 32/33a=−1m=3233 // inclination of the normal line nn

P in n: y = ax + bP∈n:y=ax+b // let's determine b

-63/16 = 32/33 cdot 4 + b−6316=3233⋅4+b

b = -63/16 - 128/33 = frac{-63*33 - 128*16}{16*33} = -4127/528b=−6316−12833=−63⋅33−128⋅1616⋅33=−4127528

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