What is the equation of the tangent line of f(x)=1x−x+1 at x=3?

Question

Sterling Mcdaniel · Accepted Answer

Explanation:First we begin by finding the value of f(3) so that we have the point of tangency locked down:f(3)=13−3+1=13−4=13−2=−53Next, in order to find the slope of the tangent at the point (3,−53) we have to have the derivative f&#039;(x) so that we can evaluate it at x=3:f(x)=x−1−(x+1)12f′(x)=(−1)(x−2)−(12)(x+1)−12=−1x2−12(x+1)12=−1x2−12x+1For x=3:f′(3)=−1(32)−123+1=−19−12⋅2=−19−14=−1336To find the equation of the tangent line, we can use the Point-Slope form of a line:y−y1=m(x−x1)y−(−53)=(−1336)(x−3)y+53=−1336x+1312y=−1336x+1312−53y=−1336x−712

liiipstick0j2 · Accepted Answer

f(x)=1x−x+1;x=3,f(x)=13−3+1∴f(x)=13−2=−53∴ The coordinates of point is(3,−53) where tangent line is drawn. Slope of the curvef(x)=1x−x+1 is f′(x)=−1x2+12x+1slope at point x=3 is m=−132+123+1 orm=−19+14 or m=536. The equation of tangent linehaving slope of m=536 paasing through point (3,−53) isy+53=536(x−3) or 36y+60=5x-15 or5x-36y=75

What is the equation of the tangent line of f(x)=\frac{1}{x}-\sqrt{x+1}

Answered question

Answer & Explanation

New Questions in Calculus 1