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Find the equation of the tangent line to

y= 1/x2+1 at x=2.

A) y = -3/5x + 11/5
B) y = -4/7x + 9/7
C) y = -8/5x+8/17
D) y= -11/x + 11/25



1 Answer

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Final answer:

The equation of the tangent line to y = 1/(x^2 + 1) at x = 2 is y = -4/25x + 11/25.

Step-by-step explanation:

To find the equation of the tangent line to the curve y = 1/(x^2 + 1) at x = 2, we need to find the derivative of the curve and evaluate it at x = 2. The derivative of y = 1/(x^2 + 1) is y' = -2x/(x^2 + 1)^2.

Substituting x = 2 into the derivative gives y'(2) = -2(2)/(2^2 + 1)^2 = -4/25.

Using the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we can substitute x = 2, y = 1/(2^2 + 1) = 1/5, and m = -4/25 to find the equation of the tangent line:

y - 1/5 = -4/25(x - 2)

Simplifying the equation gives:

y = -4/25x + 11/25

User Nikita Jajodia
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