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Question 5

The linear approximation at [tex]$x=0$[/tex] to [tex]$\frac{1}{\sqrt{2-x}}$[/tex] is [tex][tex]$A + Bx$[/tex][/tex]. What is the value of [tex]$A$[/tex]?

[tex]\square[/tex]

Sagot :

GinnyAnswer
To find the linear approximation [tex]\( A + Bx \)[/tex] of the function [tex]\( \frac{1}{\sqrt{2-x}} \)[/tex] at [tex]\( x=0 \)[/tex], we follow these steps:

1. Identify the function and the point of approximation:
Given function: [tex]\( f(x) = \frac{1}{\sqrt{2-x}} \)[/tex]
Point of approximation: [tex]\( x = 0 \)[/tex]

2. Calculate [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{\sqrt{2-0}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} = 0.7071067811865475 \][/tex]
This is the value of [tex]\( A \)[/tex], as [tex]\( A = f(0) \)[/tex].

So, the value of [tex]\( A \)[/tex] in the linear approximation at [tex]\( x = 0 \)[/tex] is [tex]\( 0.7071067811865475 \)[/tex].
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