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Use linear approximation, i.e. the tangent line, to approximate √ 16.2 as follows: Let f ( x ) = √ x . Find the equation of the tangent line to f ( x ) at x = 16 L ( x ) = Using this, we find our approximation for √ 16.2 is NOTE: For this part, give your answer to at least 9 significant figures or use an expression to give the exact answer. Get help: Video

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LammettHash

The linear approximation to [tex]f(x)[/tex] near [tex]x=16[/tex] is

[tex]L(x)=f(16)+f'(16)(x-16)[/tex]

We have

[tex]f(x)=\sqrt x\implies f(16)=4[/tex]

[tex]f'(x)=\dfrac1{2\sqrt x}\implies f'(16)=\dfrac18[/tex]

[tex]\implies L(x)=4+\dfrac18(x-16)=\dfrac x8+2[/tex]

Then

[tex]f(16.2)\approx L(16.2)\iff\sqrt{16.2}\approx\dfrac{16.2}8+2[/tex]

[tex]\implies\boxed{\sqrt{16.2}\approx4.025=\dfrac{161}{40}}[/tex]

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