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]
