Find the linear approximation of the function f(x) = √4-x at a = 0 and use it to approximate the numbers √3.9 and √3.99 ? (Round your answers to four decimal places.)

Answer 1

Use the Taylor form:

#f(x)=f(a)+(x-a)*f'(a)....#
#f(x)'=(sqrt(4-x))'=1/2(-1)/sqrt(4-x)#
#f(0)=sqrt(4-0)=2#
#f(0)'=1/2(-1)/sqrt(4)=-1/4#
#sqrt(3.9)=f(x)~~ 2+0.1(-1/4)=1.9750#
#sqrt(3.99)=f(x)~~ 2+0.01(-1/4)=1.9975#
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Answer 2

The linear approximation of the function ( f(x) = \sqrt{4 - x} ) at ( a = 0 ) is given by ( L(x) = f(a) + f'(a)(x - a) ). First, find the derivative of ( f(x) ): ( f'(x) = -\frac{1}{2\sqrt{4 - x}} ). Then, evaluate ( f(0) ) and ( f'(0) ): ( f(0) = \sqrt{4 - 0} = 2 ) and ( f'(0) = -\frac{1}{2\sqrt{4 - 0}} = -\frac{1}{4} ). Substituting these values into the linear approximation formula gives ( L(x) = 2 - \frac{1}{4}x ).

Now, to approximate ( \sqrt{3.9} ) and ( \sqrt{3.99} ) using this linear approximation:

  • For ( x = 3.9 ), ( L(3.9) = 2 - \frac{1}{4}(3.9) \approx 1.025 ).
  • For ( x = 3.99 ), ( L(3.99) = 2 - \frac{1}{4}(3.99) \approx 1.0025 ).

So, the approximations for ( \sqrt{3.9} ) and ( \sqrt{3.99} ) using the linear approximation are approximately ( 1.025 ) and ( 1.0025 ), respectively.

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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.

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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