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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]