# SCALCET8 3.10.025. Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3

SCALCET8 3.10.025. Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 126

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1. quangkhai

$$f(126) \approx 5.01333$$

Step-by-step explanation:

Given

$$\sqrt[3]{126}$$

Required

Solve using differentials

In differentiation:

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

Express 126 as 125 + 1;

i.e.

$$x = 125; \triangle x = 1$$

So, we have:

$$f(125+1) \approx f(125) + 1 \cdot f'(125)$$

$$f(126) \approx f(125) + 1 \cdot f'(125)$$

To calculate f(125), we have:

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

$$f(125) = \sqrt[3]{125}$$

$$f(125) = 5$$

So:

$$f(126) \approx f(125) + 1 \cdot f'(125)$$

$$f(126) \approx 5 + 1 \cdot f'(125)$$

$$f(126) \approx 5 + f'(125)$$

Also:

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

Rewrite as:

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

Differentiate

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

Using law of indices, we have:

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

So:

$$f'(125) = \frac{125^\frac{1}{3}}{3*125}$$

$$f'(125) = \frac{5}{375}$$

$$f'(125) = \frac{1}{75}$$

So, we have:

$$f(126) \approx 5 + f'(125)$$

$$f(126) \approx 5 + \frac{1}{75}$$

$$f(126) \approx 5 + 0.01333$$

$$f(126) \approx 5.01333$$