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How do you estimate the quantity using the Linear Approximation of #(3.9)^(1/2)#?

Answer 1

Aurora Alvarado

Function #f(x) = x^(1/2) = sqrtx#

Find an #a# near #3.9# with #f(a)# easy to find.

Use #a=4#, so #f(a) = f(4) = sqrt4 = 2#

The linear approximation of #f# at #(a,f(a))# is a form of the equation for the tangent line at #(a,f(a))#

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

For #f(x) = sqrtx#, we have #f'(x) = 1/(2sqrtx)#.

So, #f'(a) = f'(4) = 1/(2sqrt4) = 1/4#

#L(x) = 2+1/4(x-4)#

#L(3.9) = 2+0.25(-0.1) = 2-0.025 = 1.975#

(Note that: #1.975^2 = 3.900625# an

#sqrt3.9 ~~ 1.974842#.)

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Answer 2

William Abdalla

To estimate the quantity using the Linear Approximation of ( (3.9)^{\frac{1}{2}} ), you first find the derivative of the function at the point you want to approximate, then use the linearization formula:

[ f(x) \approx f(a) + f'(a)(x - a) ]

Given ( f(x) = \sqrt{x} ) and ( a = 4 ), the derivative is ( f'(x) = \frac{1}{2\sqrt{x}} ). Evaluating at ( x = 4 ), we get ( f'(4) = \frac{1}{4} ).

Using the linearization formula,

[ f(x) \approx f(4) + f'(4)(x - 4) ]

[ \sqrt{x} \approx \sqrt{4} + \frac{1}{4}(x - 4) ]

[ \sqrt{x} \approx 2 + \frac{1}{4}(x - 4) ]

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Answer from HIX Tutor

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.