# How do you find the Linear Approximation at x=0 of #y=sqrt(3+3x)#?

I got:

The Linear Approximation should be a line that can substitute (in a narrow interval) your curve.

To find the equation of this line we need to find the slope

The slope can be found deriving our function and evaluating the derivative at

At

This will be the slope

The coordinate

So, the line through our point and having slope

Or

Graphically:

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The linear approximation at ( x = 0 ) of ( y = \sqrt{3 + 3x} ) is given by ( L(x) = f'(0)(x - 0) + f(0) ), where ( f(x) = \sqrt{3 + 3x} ).

First, find ( f'(x) ): [ f'(x) = \frac{d}{dx} \sqrt{3 + 3x} ] [ = \frac{1}{2\sqrt{3 + 3x}} \cdot 3 ] [ = \frac{3}{2\sqrt{3 + 3x}} ]

Next, evaluate ( f'(0) ): [ f'(0) = \frac{3}{2\sqrt{3}} ] [ = \frac{3}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} ] [ = \frac{3\sqrt{3}}{6} ] [ = \frac{\sqrt{3}}{2} ]

Now, find ( f(0) ): [ f(0) = \sqrt{3 + 3(0)} ] [ = \sqrt{3} ]

Therefore, the linear approximation at ( x = 0 ) of ( y = \sqrt{3 + 3x} ) is: [ L(x) = \frac{\sqrt{3}}{2} \cdot x + \sqrt{3} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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