Question

1. the tangent line to $f(x)=(1 + x)^{1/4}$ at $a = 0$ is $y=1+\frac{1}{4}x$. using this tangent line to approximate $(1.1)^{1/4}$ gives: (a) $(1.1)^{1/4}approx\frac{37}{36}$ (b) $(1.1)^{1/4}approx\frac{39}{40}$ (c) $(1.1)^{1/4}approx1$ (d) $(1.1)^{1/4}approx\frac{41}{40}$ (e) $(1.1)^{1/4}approx\frac{5}{4}$

Explanation:

Step1: Recall tangent - line approximation formula

The tangent - line approximation of a function $y = f(x)$ at $x = a$ is given by $L(x)=f(a)+f^{\prime}(a)(x - a)$. For $f(x)=(1 + x)^{\frac{1}{4}}$, $a = 0$, and $x=0.1$. First, find $f(0)$ and $f^{\prime}(x)$.
$f(0)=(1 + 0)^{\frac{1}{4}}=1$.
Differentiate $y=(1 + x)^{\frac{1}{4}}$ using the power - rule $(u^n)^\prime=nu^{n - 1}u^\prime$. Here $u = 1 + x$, $n=\frac{1}{4}$, so $f^{\prime}(x)=\frac{1}{4}(1 + x)^{-\frac{3}{4}}$, and $f^{\prime}(0)=\frac{1}{4}(1+0)^{-\frac{3}{4}}=\frac{1}{4}$.
The tangent - line approximation $L(x)=f(0)+f^{\prime}(0)(x - 0)=1+\frac{1}{4}x$.

Step2: Substitute $x = 0.1$ into the tangent - line formula

When $x = 0.1$, $L(0.1)=1+\frac{1}{4}(0.1)=1+\frac{1}{40}=\frac{40 + 1}{40}=\frac{41}{40}$.
So $(1.1)^{\frac{1}{4}}\approx\frac{41}{40}$.

Answer:

d. $(1.1)^{\frac{1}{4}}\approx\frac{41}{40}$