What is the equation of the line that is normal to #f(x)= lnx^2-x# at # x= 1 #?

1 Answer
Mar 20, 2016

#y=-x#

Explanation:

A normal line is simply a line perpendicular to a tangent line. We are being asked to find the normal line at #x=1#. In order to do that, we take the derivative, evaluate it at at #x=1# (which gives us the slope at #x=1#), then use that information and a point on the line to find the normal line.

Step 1: Find the Derivative
Note first that #lnx^2# can be rewritten using the properties of logs to #2lnx#. Taking the derivative now is extremely easy: the derivative of #lnx# is #1/x#, which means the derivative of #2lnx=2/x#. As for #-x#, well, the derivative of that is just #-1#. Applying this to the problem:
#f'(x)=2/x-1#
And that's all for this step.

Step 2: Evaluate
Here, we evaluate #f'(1)# to find the slope at #x=1#:
#f'(x)=2/x-1#
#f'(1)=2/(1)-1=2-1=1#
But we don't want the slope of the tangent line, we want the slope of the normal line. Luckily, there is a simple relationship between tangent line and normal line slopes: they are opposite reciprocals. That is to say:
Normal line slope=-1/tangent line slope

In our case, that means the normal line slope is #-1/1=-1#.

Step 3: Normal Line Equation
Normal lines, like tangent lines, are of the form #y=mx+b#, where #x# and #y# are points on the line, #m# is the slope, and #b# is the #y#-intercept. We have the slope (#-1#), and we can easily get two points on the line. Using #x=1#, we have:
#f(1)=2ln(1)-(1)=2(0)-1=-1#

Now we can solve for #b#:
#y=mx+b#
#-1=(1)(-1)+b#
#-1=b-1#
#b=0#

The equation of the normal line is therefore #y=-x#.