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

2 Answers
Feb 25, 2016
  • The equation of the line is:
    y=1.37x+2.7

Explanation:

  • to find the equation of the line, we find its slope (m) knowing that:
    m=f'(x)
  • f'(x)=d/dx[e^(1/x)-x]
    by substituting x with 1/z in the main equation:
    f(1/z)=e^z-1/z
    f'(x)=f'(1/z)=d/dz[e^z-1/z]
    =e^z+1/z^2 (because the derivative of e^x is e^x )
    =e^(1/x)+x^2

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

  • f'(-1)=e^(-1)+1=1.37
    :.m=1.37
  • The equation of the line is:
    y-1.37=1.37(x+1)
    y=1.37x+2.7
Mar 2, 2016

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

~~ 0.731 x + 2.099

Explanation:

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).

Self-drawnSelf-drawn