How do you find the linearization at a=16 of #f(x) = x^(1/2)#?
Use the formula
This can be used to obtain good approximations to square roots of numbers near 16. For example,
A more accurate approximation with technology is:
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To find the linearization of (f(x) = \sqrt{x}) at (a = 16), we follow these steps:

Find the first derivative of (f(x)) with respect to (x). [f'(x) = \frac{1}{2\sqrt{x}}]

Evaluate (f'(x)) at (x = a). [f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8}]

Write the equation of the tangent line using the pointslope form: [y  f(a) = f'(a)(x  a)]

Substitute the values of (a), (f(a)), and (f'(a)). [y  f(16) = \frac{1}{8}(x  16)]

Simplify the equation to obtain the linearization. [y  4 = \frac{1}{8}(x  16)]
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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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