Question

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

Answer 1

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:

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

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

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

$f(216) \approx 6 + 0.0093$

$f(216) \approx 6.0093$