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Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 217

Sagot :

MrRoyal

Answer:

[tex]f(216) \approx 6.0093[/tex]

Step-by-step explanation:

Given

[tex]\sqrt[3]{217}[/tex]

Required

Solve

Linear approximated as:

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

Take:

[tex]x = 216; \triangle x= 1[/tex]

So:

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

Substitute 216 for x

[tex]f(x) = \sqrt[3]{216}[/tex]

[tex]f(x) = 6[/tex]

So, we have:

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

[tex]f(215 + 1) \approx 6 +1 \cdot f'(x)[/tex]

[tex]f(216) \approx 6 +1 \cdot f'(x)[/tex]

To calculate f'(x);

We have:

[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]

Split

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

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

Substitute 216 for x

[tex]f'(216) = \frac{216^\frac{1}{3}}{3*216}[/tex]

[tex]f'(216) = \frac{6}{648}[/tex]

[tex]f'(216) = \frac{3}{324}[/tex]

So:

[tex]f(216) \approx 6 +1 \cdot f'(x)[/tex]

[tex]f(216) \approx 6 +1 \cdot \frac{3}{324}[/tex]

[tex]f(216) \approx 6 + \frac{3}{324}[/tex]

[tex]f(216) \approx 6 + 0.0093[/tex]

[tex]f(216) \approx 6.0093[/tex]

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