What is the equation of the tangent line of f(x)=x+1−x+2 at x=2?

f(x)=x+1−x+2f′(x)=12x+1−12x+2f′(x)=(12)x+2−x+1x+1x+2At (2,3−2)Sub x=2 into f'(x)f′(2)=(12)2+2−2+12+12+2f′(2)=2−343Equation of the tangent:(y−3+2)=2−343(x−2)4y3−12+83=2x−4−x3+234y3=x(2−3)+8−63

Question: "What is the equation of the tangent line..."

High School
Calculus and Analysis
Calculus 1
Derivatives

amonitas3zeb

Answered question

2022-03-22

What is the equation of the tangent line of

f (x) = \sqrt{x + 1} - \sqrt{x + 2}

at x=2?

Answer & Explanation

Harry Gibson

Beginner2022-03-23Added 13 answers

f (x) = \sqrt{x + 1} - \sqrt{x + 2}

f^{'} (x) = \frac{1}{2 \sqrt{x + 1}} - \frac{1}{2 \sqrt{x + 2}}

f^{'} (x) = (\frac{1}{2}) \frac{\sqrt{x + 2} - \sqrt{x + 1}}{\sqrt{x + 1} \sqrt{x + 2}}

(2, \sqrt{3} - 2)

Sub x=2 into f'(x)

f^{'} (2) = (\frac{1}{2}) \frac{\sqrt{2 + 2} - \sqrt{2 + 1}}{\sqrt{2 + 1} \sqrt{2 + 2}}

f^{'} (2) = \frac{2 - \sqrt{3}}{4 \sqrt{3}}

Equation of the tangent:

(y - \sqrt{3} + 2) = \frac{2 - \sqrt{3}}{4 \sqrt{3}} (x - 2)

4 y \sqrt{3} - 12 + 8 \sqrt{3} = 2 x - 4 - x \sqrt{3} + 2 \sqrt{3}

4 y \sqrt{3} = x (2 - \sqrt{3}) + 8 - 6 \sqrt{3}

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