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Olenka [21]
3 years ago
12

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

Mathematics
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

f(216) \approx 6.0093

Step-by-step explanation:

Given

\sqrt[3]{217}

Required

Solve

Linear approximated as:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

Take:

x = 216; \triangle x= 1

So:

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

Substitute 216 for x

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

f(x) = 6

So, we have:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

f(215 + 1) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot f'(x)

To calculate f'(x);

We have:

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

Rewrite as:

f(x) = x^\frac{1}{3}

Differentiate

f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}

Split

f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}

f'(x) = \frac{x^\frac{1}{3}}{3x}

Substitute 216 for x

f'(216) = \frac{216^\frac{1}{3}}{3*216}

f'(216) = \frac{6}{648}

f'(216) = \frac{3}{324}

So:

f(216) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot \frac{3}{324}

f(216) \approx 6  + \frac{3}{324}

f(216) \approx 6  + 0.0093

f(216) \approx 6.0093

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