Question

find the linearization ( l(x) ) of the function ( f(x)=e^{2 x} ) at ( x = 0 ).
answer: ( l(x)= )

Explanation:

Step1: Recall the formula for linearization

The formula for the linearization \(L(x)\) of a function \(y = f(x)\) at \(x = a\) is \(L(x)=f(a)+f^{\prime}(a)(x - a)\). Here \(a = 0\), \(f(x)=e^{2x}\).

Step2: Find \(f(0)\)

Substitute \(x = 0\) into \(f(x)\): \(f(0)=e^{2\times0}=e^{0}=1\).

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

Using the chain - rule, if \(y = e^{u}\) and \(u = 2x\), then \(y^{\prime}=\frac{dy}{du}\cdot\frac{du}{dx}\). Since \(\frac{dy}{du}=e^{u}\) and \(\frac{du}{dx}=2\), \(f^{\prime}(x)=2e^{2x}\).

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

Substitute \(x = 0\) into \(f^{\prime}(x)\): \(f^{\prime}(0)=2e^{2\times0}=2\times1 = 2\).

Step5: Substitute into the linearization formula

Since \(a = 0\), \(L(x)=f(0)+f^{\prime}(0)(x - 0)\). Substituting \(f(0)=1\) and \(f^{\prime}(0)=2\) into the formula, we get \(L(x)=1 + 2x\).

Answer:

\(L(x)=1 + 2x\)