Question

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

Answer 1

Answer:

$f(126) \approx 5.01333$

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:

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

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

$f(126) \approx 5 + 0.01333$

$f(126) \approx 5.01333$