What is the equation of the tangent line of #f(x)=1/(2x)+4 # at #x=2#?

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Answer 1 · 2015-12-01T15:54:26

#y=1/8x+4#

The slope of the tangent line at the point #(2,17/4)# is #f'(2)#.

To find the derivative, the easiest way is to rewrite #f(x)# as such:

#f(x)=1/2x^-1+4#

Recall that #d/dx[x^n]=nx^(n-1)#.

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

Thus, #f'(2)=-1/(2((-2)^2))=-1/8#

In point-slope form: #y-17/4=-1/8(x-2)#

In slope-intercept form: #y=-1/8x+9/2#