Respuesta :

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[tex]f(x)=\sqrt[3]{1+x}=(1+x)^{1/3}\implies f'(x)=\dfrac1{3(1+x)^{2/3}}[/tex]

The linear approximation to [tex]f(x)[/tex] around [tex]x=a[/tex] is then

[tex]L(x)=f(a)+f'(a)(x-a)\approx f(x)[/tex]

So the approximation centered at [tex]a=0[/tex] will be

[tex]L(x)=f(0)+f'(0)x=(1+0)^{1/3}+\dfrac x{3(1+0)^{2/3}}[/tex]
[tex]L(x)=1+\dfrac x3[/tex]

which means we have

[tex]\sqrt[3]{0.96}\approx L(0.96)=1+\dfrac{0.96}3=1.32[/tex]

[tex]\sqrt[3]{1.02}\approx L(1.02)=1+\dfrac{1.02}3=1.34[/tex]

Compare to the actual values of

[tex]\sqrt[3]{0.96}\approx0.9864[/tex]

[tex]\sqrt[3]{1.02}\approx1.0066[/tex]

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