For #f(x)=1/x# what is the equation of the tangent line at #x=1/2#?

1 Answer
Nov 21, 2015

#y=-4x+4#

Explanation:

We must find #f'(1/2)# to determine the slope of the tangent line.

Remember that the derivative of #x^n# is #nx^(n-1)#.

#f(x)=x^-1#
#f'(x)=-x^(-2)#
#f'(x)=-1/x^2#

#f'(1/2)=-1/(1/2)^2=-1/(1/4)=-4#

The slope of our tangent line at #x=1/2# is #-4#. We know that the point the tangent line touches is #(1/2,2)# by finding #f(1/2)#.

We can use point-slope form to write our linear equation:
#y-2=-4(x-1/2)#

If you want your answer in slope-intercept form:
#y=-4x+4#