How do you find the linearization at a=3 of # f(x) =sqrt(x² + 2)#?
The linearization of
The linearization/tangent line can be written:
Or, in slope intercept for as
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To find the linearization of ( f(x) = \sqrt{x^2 + 2} ) at ( a = 3 ), follow these steps:

Find the derivative of ( f(x) ). [ f'(x) = \frac{1}{2\sqrt{x^2 + 2}} \cdot 2x = \frac{x}{\sqrt{x^2 + 2}} ]

Evaluate ( f'(a) ) at ( a = 3 ). [ f'(3) = \frac{3}{\sqrt{3^2 + 2}} = \frac{3}{\sqrt{11}} ]

Find the equation of the tangent line using the pointslope form ( y  y_1 = m(x  x_1) ), where ( (x_1, y_1) ) is the point ( (a, f(a)) ) and ( m ) is the slope ( f'(a) ). [ y  f(3) = f'(3)(x  3) ]

Substitute ( f(3) ) and ( f'(3) ). [ y  \sqrt{3^2 + 2} = \frac{3}{\sqrt{11}}(x  3) ]

Simplify the equation to obtain the linearization.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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