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 substitutingx with1/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 ofe^x ise^x )
=e^(1/x)+x^2
At
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
~~ 0.731 x + 2.099
Explanation:
The normal line passes through the point
f(-1) = 1/e + 1
The normal line is perpendicular to the tangent line. It has a gradient of
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
Self-drawn