Question

linearization and differentials: problem 3
(1 point)
find the linearization ( l(x) ) of the function ( f(x)=e^{-5 x} ) at ( x = 0 ).
answer: ( l(x)= )
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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 \(a = 0\), \(f(x)=e^{-5x}\).

Step2: Find \(f(0)\)

Substitute \(x = 0\) into \(f(x)\): \(f(0)=e^{-5\times0}=e^{0}=1\).

Step3: Find the derivative \(f^{\prime}(x)\)

Using the chain - rule, if \(y = e^{u}\) and \(u=-5x\), then \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). Since \(\frac{dy}{du}=e^{u}\) and \(\frac{du}{dx}=-5\), \(f^{\prime}(x)=e^{-5x}\times(-5)=-5e^{-5x}\).

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

Substitute \(x = 0\) into \(f^{\prime}(x)\): \(f^{\prime}(0)=-5e^{-5\times0}=-5\).

Step5: Substitute into the linearization formula

Substitute \(a = 0\), \(f(0)=1\) and \(f^{\prime}(0)=-5\) into \(L(x)=f(a)+f^{\prime}(a)(x - a)\). We get \(L(x)=1+(-5)(x - 0)\).

Answer:

\(L(x)=1 - 5x\)