3. -/4 points details my notes scalc9 2.9.006.mi. find the linear approximation l(x) of the function g(x) =…

Explanation:

Step1: Recall linear - approximation formula

The linear approximation of a function $y = g(x)$ at $x = a$ is given by $L(x)=g(a)+g^{\prime}(a)(x - a)$. First, find $g(x)=(1 + x)^{\frac{1}{3}}$, then $g(a)$ and $g^{\prime}(x)$.

Step2: Calculate $g(a)$ at $a = 0$

Substitute $x=a = 0$ into $g(x)$. So, $g(0)=(1+0)^{\frac{1}{3}}=1$.

Step3: Find the derivative of $g(x)$

Using the power - rule $(u^n)^\prime=nu^{n - 1}u^\prime$, if $u = 1 + x$ and $n=\frac{1}{3}$, then $g^{\prime}(x)=\frac{1}{3}(1 + x)^{-\frac{2}{3}}$.

Step4: Calculate $g^{\prime}(a)$ at $a = 0$

Substitute $x = 0$ into $g^{\prime}(x)$. So, $g^{\prime}(0)=\frac{1}{3}(1+0)^{-\frac{2}{3}}=\frac{1}{3}$.

Step5: Find the linear approximation $L(x)$

Substitute $g(0) = 1$ and $g^{\prime}(0)=\frac{1}{3}$ into the linear - approximation formula $L(x)=g(a)+g^{\prime}(a)(x - a)$. We get $L(x)=1+\frac{1}{3}x$.

Step6: Approximate $\sqrt[3]{0.95}$

Rewrite $\sqrt[3]{0.95}=\sqrt[3]{1+( - 0.05)}$. Let $x=-0.05$ in $L(x)$. Then $L(-0.05)=1+\frac{1}{3}(-0.05)=1-\frac{0.05}{3}\approx0.983$.

Step7: Approximate $\sqrt[3]{1.1}$

Rewrite $\sqrt[3]{1.1}=\sqrt[3]{1 + 0.1}$. Let $x = 0.1$ in $L(x)$. Then $L(0.1)=1+\frac{1}{3}(0.1)=1+\frac{0.1}{3}\approx1.033$.

Answer:

$L(x)=1+\frac{1}{3}x$; $\sqrt[3]{0.95}\approx0.983$; $\sqrt[3]{1.1}\approx1.033$