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

Question

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

caggeosteanc · Accepted Answer

The linearization of a differentiable function

f

at a point

x = a

is the linear function

L (x) = f (a) + f' (a) (x - a)

, whose graph is the tangent line to the graph of

f

at the point

(a, f (a))

. When

x \approx a

, we get the approximation

f (x) \approx L (x)

.
For

f (x) = \sqrt{x + 3} = {(x + 3)}^{\frac{1}{2}}

we get

f' (x) = \frac{1}{2} \cdot {(x + 3)}^{- \frac{1}{2}}

so that

f (1) = \sqrt{4} = 2

and

f' (1) = \frac{1}{2} \cdot 4^{- \frac{1}{2}} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

. Therefore,

f (x) \approx L (x) = 2 + (\frac{1}{4}) (x - 1) = \frac{x}{4} + \frac{7}{4}

How to find the linearization at a=1 of f(x) =sqrt(x+3)?

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