linearization and differentials: problem 3\n(1 point)\nfind the linearization ( l(x) ) of the function (…

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)