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

Question

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

William Abdalla · Accepted Answer

#y=-x# 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#.

Emily Ammerman · Answer

undefined The equation of the line that is normal to f(x) = lnx^2 - x at x = 1 is y = -2x + 1.