선형근사와 미분소

We approximate the function $f(x)=\sqrt[5]{x}$. We need to centre the approximation about some value $x=a$ such that we know $f(a)$ and $f'(a)$, and $a$ is not too far from $30$.

$$\begin{aligned} f(x)&=\sqrt[5]{x}=x^{\frac{1}{5}}\\ f'(x)&=\frac{1}{5}x^{-\frac{4}{5}}=\frac{1}{5\sqrt[5]{x}^4} \\ \text{$a$ needs to be a number whose fifth root we know. Since $\sqrt[5]{32}=2$, and 32 is reasonably close to 30, $a=32$ is a great choice.} \\ f(32)&=\sqrt[5]{32}=2\\ f'(32)&=\frac{1}{5\cdot 2^4}=\frac{1}{80} \\ \text{The linear approximation of $f(x)$ about $x=32$ is} \\ f(x) &\approx 2+\frac{1}{80}(x-32) \\ \text{When $x=30$:} \\ f(30) &\approx 2+\frac{1}{80}(30-32)=2-\frac{1}{40}=\frac{79}{40} \end{aligned}$$

We estimate $\sqrt[5]{30}\approx \dfrac{79}{40}$.

Remark: $\dfrac{79}{40} = 1.975$, while $\sqrt[5]{30} \approx 1.97435$. This is a decent estimation.