What is the equation of the line tangent to #f(x)=1/(5+4x)^2 # at #x=7#?

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Answer 1 · 2016-02-05T01:00:33

In point-slope form, it is #y-1/33^2 = -8/33^3(x-7)#

At #x=7#, we get #y=1/33^3#, so the tangent goes through the point #(7,1/33^2)#

#f(x) = (5+4x)^-2#, so

#f'(x) = -2(5+4x)^-3 (4) = (-8)/(5+4x)^3#.

The slope of the tangent at #x=7# is #f'(7) = -8/33^3#.

Finish by writing the equation

#y-1/33^2 = -8/33^3(x-7)#.

If you need slope-intercept form, do the algebra to solve for #y=-8/33^3x+89/33^3#.