Asked by Sandra Gibson
Find an equation of the tangent line to the graph of y = f(x) at x = a, a = 1.
f(x) = x + 3/x^2 + 1
f(x) = x + 3/x^2 + 1
Answers
Answered by
bobpursley
I assume you are a calculus student.
slope tangent= f'= 1/(x^2+1)-x/(x^2+1)^2 -3/(x^2+1)^2
put in x=a, and you have slope m
then y=mx+b
put in x=a, on the equation of f(x), solve for y, then here solve for b.
then you have it.
slope tangent= f'= 1/(x^2+1)-x/(x^2+1)^2 -3/(x^2+1)^2
put in x=a, and you have slope m
then y=mx+b
put in x=a, on the equation of f(x), solve for y, then here solve for b.
then you have it.
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