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

Answer 1

Scarlett Allison

The equation of the line is:
#y=1.37x+2.7#

At #x= -1# #y=f(-1)=e^(-1)+1=1.37#

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Answer 2

Aurora Alvarado

#y = frac{e}{e+1}x + 2 + frac{1]{e(e+1)}#

#~~ 0.731 x + 2.099#

The normal line passes through the point #(-1,f(-1))#.

#f(-1) = 1/e + 1#

The normal line is perpendicular to the tangent line. It has a gradient of #m=frac{-1}{f'(-1)}#.

#f'(x) = frac{"d"}{"d"x}(e^{1/x}) - frac{"d"}{"d"x}(x)#

#= e^{1/x}frac{"d"}{"d"x}(1/x) - 1#
#= -1/x^2 e^(1/x) - 1#

#f'(-1) = -1/e -1#
#m = frac{-1}{-1/e -1}#

#= frac{e}{e + 1}#
Now we know the gradient and a point, we can write the equation of the normal line in point-slope form.

#y-f(-1) = m(x-(-1))#
#y-(1/e + 1) = frac{e}{e + 1}(x+1)#

Rewrite it in slope-intercept form.

#y = frac{e}{e+1}x + 2 + frac{1]{e(e+1)}#

Here is a graph of #y=f(x)#.

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Answer 3

Christopher Allers

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

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Answer from HIX Tutor

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