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Use linear approximation to approximate √25.3 as follows.
Let f(x)=√x. The equation of the tangent line to f(x) at x=25 can be written in the form y=mx+b. Compute m,b.
Using this, find the approximation for √25.3.

Respuesta :

LammettHash

The idea is to use the tangent line to [tex]f(x)=\sqrt x[/tex] at [tex]x=25[/tex] in order to approximate [tex]f(25.3)=\sqrt{25.3}[/tex].

We have

[tex]f(x)=\sqrt x\implies f(25)=\sqrt{25}=5[/tex]

[tex]f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}[/tex]

so the linear approximation to [tex]f(x)[/tex] is

[tex]L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92[/tex]

Hence [tex]m=\frac1{10}[/tex] and [tex]b=\frac92[/tex].

Then

[tex]f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}[/tex]

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