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

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Answer 1 · 2016-01-03T22:22:45

#y=x+2#

Rewrite #f(x)#:

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

Use the chain rule to differentiate #f(x)#:

#d/dx(u^-1)=-u^-2=-1/u^2#

Thus,

#f'(x)=-1/(x^2+x+1)^2*d/dx(x^2+x+1)#

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

To find the slope of the tangent line, calculate #f'(-1)#.

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

Find the point the tangent line will intersect:

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

The tangent line will intersect the point #(-1,1)# and have slope #1#.

Write in point slope form:

#y-1=1(x+1)#

Or, in slope intercept form:

#y=x+2#

graph{(1/(x^2+x+1)-y)(x+2-y)=0 [-10.08, 9.92, -3.12, 6.88]}