answer: \\frac{\\sqrt{2}+5}{2} \n# ____ f(x)=\\frac{1}{x} f(e)=-3 \nto advance in the circuit, locate…

answer: \\frac{\\sqrt{2}+5}{2} \n# ____ f(x)=\\frac{1}{x} f(e)=-3 \nto advance in the circuit, locate f(e^{4}).

answer: \\frac{\\sqrt{2}+5}{2} \n# ____ f(x)=\\frac{1}{x} f(e)=-3 \nto advance in the circuit, locate f(e^{4}).

Answer

Explanation:

Step1: Integrate (f^{\prime}(x))

Since (f^{\prime}(x)=\frac{1}{x}), then (f(x)=\int\frac{1}{x}dx=\ln|x| + C) (where (C) is the constant of integration).

Step2: Find the value of (C)

Given (f(e)=-3), substitute (x = e) into (f(x)=\ln|x|+C). So (\ln(e)+C=-3). Since (\ln(e) = 1), we have (1 + C=-3), then (C=-4).

Step3: Find (f(e^{4}))

Substitute (x = e^{4}) into (f(x)=\ln|x|-4). Then (f(e^{4})=\ln(e^{4})-4). Using the property (\ln(a^{b})=b\ln(a)), we get (f(e^{4})=4\ln(e)-4). Since (\ln(e) = 1), (f(e^{4})=4 - 4=0).

Answer:

(0)