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

Answer 1

Isaiah Allen

#y = 32/33 x - 4127/528 Leftrightarrow 528y = 512x - 4127#

#f(4) = 1/16 - 4 Rightarrow P = (4, -63/16)#

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

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

#a = -1/m = 32/33# // inclination of the normal line #n#

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

#-63/16 = 32/33 cdot 4 + b#

#b = -63/16 - 128/33 = frac{-63*33 - 128*16}{16*33} = -4127/528#

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Answer 2

Charles Arocha

The equation of the normal line of f(x)=1/x^2-x at x = 4 is y = -3x + 13.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.