Respuesta :

DivineSolar
So let f(x) = x^(2/3) 
Then let f'(x) = 2/3 x^(-1/3) = 2 / (3x^(1/3)) 

When x = 8, 
f(8) = 8^(2/3) = 4 
f'(8) = 2 / (3*8^(1/3)) = 1/3 

So near x = 8, the linear approximation is 
f(x) ≈ f(8) + f'(8) (x - 8) 
f(x) ≈ 4 + 1/3 (x - 8) 

So the linear approximation for x = 8.03 is... 
f(8.03) ≈ 4 + 1/3 (8.03 - 8) 
f(8.03) ≈ 4 + 1/3 (0.03) 
f(8.03) ≈ 4.01 

8.03^(2/3) ≈ 4.01 
dfrincon99

Answer:

[tex]8.07^{2/3}[/tex] ≈ 4.0233...

Step-by-step explanation:

First we identify the function: [tex]f(x)=\sqrt[3]{x^{2} }[/tex]

then we take the firts derivative: [tex]f(x)=\frac{2}{3\sqrt[3]{x}}[/tex]

Then we take a starting point a=8, so the function has the value:

[tex]f(8)=\sqrt[3]{8^{2} }=4[/tex]

and the first derivative has the value:

[tex]f(8)=\frac{2}{3\sqrt[3]{8}}=\frac{1}{3}[/tex]

Then consider the folowing relation:

f(x) ≈ f(a) + f'(a) (Δx); where Δx = x-a = 8.07-8 = 0.07

 Finally we replace the values and find:

[tex]8.07^{2/3}[/tex] ≈ 4 +( [tex]\frac{1}{3}[/tex] * 0.07)

[tex]8.07^{2/3}[/tex] ≈ 4.0233...

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