find the linearization $l(x)$ of the function $f(x)=e^{4x}$ at $x = 0$. answer: $l(x)=$

Explanation:

Step1: Recall linearization formula

The linearization $L(x)$ of a function $y = f(x)$ at $x = a$ is given by $L(x)=f(a)+f^{\prime}(a)(x - a)$.

Step2: Find $f(0)$

Given $f(x)=e^{4x}$, then $f(0)=e^{4\times0}=e^{0}=1$.

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

Using the chain - rule, if $y = e^{u}$ and $u = 4x$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Since $\frac{d}{du}(e^{u})=e^{u}$ and $\frac{d}{dx}(4x)=4$, we have $f^{\prime}(x)=4e^{4x}$.

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

Substitute $x = 0$ into $f^{\prime}(x)$, so $f^{\prime}(0)=4e^{4\times0}=4$.

Step5: Calculate $L(x)$

Substitute $a = 0$, $f(0)=1$ and $f^{\prime}(0)=4$ into the linearization formula $L(x)=f(a)+f^{\prime}(a)(x - a)$. We get $L(x)=1 + 4(x-0)=1 + 4x$.

Answer:

$1 + 4x$