# Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 9

$\sqrt[4] {1+2x} \approx 1+\frac{1}{2}~x$ On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.3 \leq x \leq 0.6$

#### Work Step by Step

$f(x) = \sqrt[4] {1+2x}$ $f'(x) = \frac{1}{4}~(1+2x)^{-3/4}\cdot (2)$ $f'(x) = \frac{1}{2~(1+2x)^{3/4}}$ When $x = 0$: $f'(x) = \frac{1}{2~(1+2x)^{3/4}}$ $f'(0) = \frac{1}{2~(1+2(0))^{3/4}}$ $f'(0) = \frac{1}{2}$ We can find the linear approximation at $a=0$: $f(x) \approx f(a)+f'(a)(x-a)$ $f(x) \approx f(0)+f'(0)(x-0)$ $f(x) \approx 1+\frac{1}{2}~x$ $\sqrt[4] {1+2x} \approx 1+\frac{1}{2}~x$ On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.3 \leq x \leq 0.6$ We can verify this: When $x = -0.3$: $1+\frac{1}{2}~x = 1+\frac{1}{2}~(-0.3) = 0.85$ $\sqrt[4] {1+2x} = \sqrt[4] {1+2(-0.3)} = 0.795$ When $x = 0.6$: $1+\frac{1}{2}~x = 1+\frac{1}{2}~(0.6) = 1.30$ $\sqrt[4] {1+2x} = \sqrt[4] {1+2(0.6)} = 1.22$

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