What is the equation of the normal line of #f(x)=ln(1/x)# at #x=5#?

Answer 1

#y=5x-25+ln(1/5)#

According to the Chain Rule,

#f'(x)=(d/dx[1/x])/(1/x)#
Let's find the derivative. (Remember that #1/x=x^-1#.)
#d/dx[x^-1]=-x^-2=-1/x^2#

Plug back in to find that:

#f'(x)=(-1/x^2)/(1/x)=-1/x^2(x/1)=-1/x#
We want the equation of the normal line at the point #(5,ln(1/5))#. The normal line is perpendicular to the tangent line, so it has an opposite reciprocal slope.
We can find the slope of the tangent line at #x=5# by calculating #f'(5)=-1/5#
Therefore, the slope of the normal line is #5#.

We can use point-slope form to write an equation:

#y-ln(1/5)=5(x-5)#

Or, in slope-intercept form:

#y=5x-25+ln(1/5)#
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Answer 2

The equation of the normal line of f(x)=ln(1/x) at x=5 is y = -5x + ln(1/5) + 5.

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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.

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.

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.

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.

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