Question

(a) find the linearization $l(x)$ of the function $sqrt{1 - 4x}+1$ at $a = 0$.
$l(x)=$ symbolic expression
(b) use part (a) to approximate the value of $1+sqrt{1.2}$.
$1+sqrt{1.2}approx$ number (3 significant figures)

Explanation:

Step1: Recall linearization formula

The linearization of a function $y = f(x)$ at $x = a$ is $L(x)=f(a)+f^{\prime}(a)(x - a)$. Let $f(x)=\sqrt{1 - 4x}+1=(1 - 4x)^{\frac{1}{2}}+1$.

Step2: Find $f(a)$ when $a = 0$

$f(0)=\sqrt{1-4\times0}+1=\sqrt{1}+1=2$.

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

Using the chain - rule, if $u = 1 - 4x$, then $y = u^{\frac{1}{2}}+1$. $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=-4$. So $f^{\prime}(x)=\frac{1}{2}(1 - 4x)^{-\frac{1}{2}}\times(-4)=-\frac{2}{\sqrt{1 - 4x}}$.

Step4: Find $f^{\prime}(a)$ when $a = 0$

$f^{\prime}(0)=-\frac{2}{\sqrt{1-4\times0}}=-2$.

Step5: Find the linearization $L(x)$

$L(x)=f(0)+f^{\prime}(0)(x - 0)=2-2x$.

Step6: Rewrite $1+\sqrt{1.2}$ to use $L(x)$

We want to approximate $1+\sqrt{1.2}$. Let $1 - 4x=1.2$, then $-4x=0.2$ and $x=-0.05$.

Step7: Use $L(x)$ for approximation

Substitute $x=-0.05$ into $L(x)$. $L(-0.05)=2-2\times(-0.05)=2 + 0.1=2.1$.

Answer:

(a) $L(x)=2 - 2x$
(b) $2.1$