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How do you find the linearization at a=1 of f(x)=√x+3?

User Frank Eno
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Answer:

y = (1/4)x + 7/4

Explanation:

To find the linearization of f(x) = √(x+3) at a=1, we need to follow these steps:

Find the first derivative of f(x) with respect to x:

f'(x) = 1 / (2√(x+3))

Evaluate f(1) to find the y-coordinate of the point where we want to find the linearization:

f(1) = √4 = 2

Evaluate f'(1) to find the slope of the tangent line at the point (1, f(1)):

f'(1) = 1 / (2√4) = 1/4

Use the point-slope form of the equation of a line to write the equation of the tangent line at (1, 2):

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

Simplify the equation of the tangent line:

y = (1/4)x + 7/4

This is the linearization of f(x) = √(x+3) at a=1.

User Gianlucaparadise
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