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Given #f(x)=sqrtx# when x=25, how do you find the linear approximation for #sqrt25.4#?

Answer 1

Luke Alvarez

#sqrt(25.4) ~~ 5.04#

Given:

#f(x) = sqrt(x) = x^(1/2)#

We have:

#f(25) = sqrt(25) = 5#

#f'(x) = 1/2 x^(-1/2)#

#f'(25) = 1/2 1/sqrt(25) = 1/10#

So:

#f(25.4) = f(25+0.4) ~~ f(25) + 0.4*f'(25) = 5+0.4/10 = 5.04#

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Answer 2

Ezra Adams

To find the linear approximation for ( \sqrt{25.4} ) using the function ( f(x) = \sqrt{x} ) when ( x = 25 ), follow these steps:

Start with the given function ( f(x) = \sqrt{x} ).
Find the derivative of the function ( f(x) ), denoted as ( f'(x) ).
Evaluate ( f(x) ) and ( f'(x) ) at ( x = 25 ) to get the values ( f(25) ) and ( f'(25) ).
Use the linear approximation formula:

[ f(a + \Delta x) ≈ f(a) + f'(a) \cdot \Delta x ]

where ( a ) is the known value (in this case, 25), ( f(a) ) is the known square root of that value, and ( \Delta x ) is the difference between the value you're estimating and the known value (in this case, ( \Delta x = 25.4 - 25 = 0.4 )).

Plug in the values ( f(25) ), ( f'(25) ), and ( \Delta x = 0.4 ) into the linear approximation formula to find the approximation for ( \sqrt{25.4} ).

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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.