How do you find the tangent line approximation for #f(x)=sqrt(1+x)# near #x=0# ?
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To find the tangent line approximation for ( f(x) = \sqrt{1 + x} ) near ( x = 0 ), follow these steps:

Find the first derivative of ( f(x) ) using the power rule or chain rule. [ f'(x) = \frac{d}{dx}\sqrt{1+x} = \frac{1}{2\sqrt{1+x}} ]

Evaluate ( f'(x) ) at ( x = 0 ) to find the slope of the tangent line at that point. [ f'(0) = \frac{1}{2\sqrt{1+0}} = \frac{1}{2} ]

Use the pointslope form of a line equation, ( y  y_1 = m(x  x_1) ), where ( (x_1, y_1) ) is the point of tangency and ( m ) is the slope of the tangent line.

Substitute ( x_1 = 0 ), ( y_1 = f(0) ), and ( m = f'(0) ) into the pointslope form. [ y  f(0) = f'(0)(x  0) ]

Simplify the equation to find the tangent line approximation. [ y  1 = \frac{1}{2}x ]
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To find the tangent line approximation for (f(x) = \sqrt{1 + x}) near (x = 0), follow these steps:
 Find the first derivative of (f(x)) with respect to (x), denoted as (f'(x)).
 Evaluate (f'(0)) to find the slope of the tangent line.
 Use the pointslope form of a line to write the equation of the tangent line.
Here's how to do it:

Find the first derivative of (f(x)): [f'(x) = \frac{d}{dx}\sqrt{1 + x}]

Evaluate (f'(0)): [f'(0) = \frac{d}{dx}\sqrt{1 + x}\bigg_{x=0}]

Use the pointslope form of a line to write the equation of the tangent line: [y  f(0) = f'(0)(x  0)]

Simplify the equation to obtain the tangent line approximation.
Following these steps:

The first derivative of (f(x) = \sqrt{1 + x}) is: [f'(x) = \frac{1}{2\sqrt{1 + x}}]

Evaluate (f'(0)): [f'(0) = \frac{1}{2\sqrt{1 + 0}} = \frac{1}{2}]

The pointslope form of the tangent line is: [y  f(0) = f'(0)(x  0)] [y  \sqrt{1 + 0} = \frac{1}{2}(x  0)] [y  1 = \frac{1}{2}x]
Therefore, the tangent line approximation for (f(x) = \sqrt{1 + x}) near (x = 0) is: [y = \frac{1}{2}x + 1]
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