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SCALCET8 3.10.025. Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 126

Sagot :

MrRoyalidn

Answer:

[tex]f(126) \approx 5.01333[/tex]

Step-by-step explanation:

Given

[tex]\sqrt[3]{126}[/tex]

Required

Solve using differentials

In differentiation:

[tex]f(x+\triangle x) \approx f(x) + \triangle x \cdot f'(x)[/tex]

Express 126 as 125 + 1;

i.e.

[tex]x = 125; \triangle x = 1[/tex]

So, we have:

[tex]f(125+1) \approx f(125) + 1 \cdot f'(125)[/tex]

[tex]f(126) \approx f(125) + 1 \cdot f'(125)[/tex]

To calculate f(125), we have:

[tex]f(x) = \sqrt[3]{x}[/tex]

[tex]f(125) = \sqrt[3]{125}[/tex]

[tex]f(125) = 5[/tex]

So:

[tex]f(126) \approx f(125) + 1 \cdot f'(125)[/tex]

[tex]f(126) \approx 5 + 1 \cdot f'(125)[/tex]

[tex]f(126) \approx 5 + f'(125)[/tex]

Also:

[tex]f(x) = \sqrt[3]{x}[/tex]

Rewrite as:

[tex]f(x) = x^\frac{1}{3}[/tex]

Differentiate

[tex]f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}\\[/tex]

Using law of indices, we have:

[tex]f'(x) = \frac{x^\frac{1}{3}}{3x}[/tex]

So:

[tex]f'(125) = \frac{125^\frac{1}{3}}{3*125}[/tex]

[tex]f'(125) = \frac{5}{375}[/tex]

[tex]f'(125) = \frac{1}{75}[/tex]

So, we have:

[tex]f(126) \approx 5 + f'(125)[/tex]

[tex]f(126) \approx 5 + \frac{1}{75}[/tex]

[tex]f(126) \approx 5 + 0.01333[/tex]

[tex]f(126) \approx 5.01333[/tex]

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