Question

answer: \frac{\sqrt{2}+5}{2}

____ f(x)=\frac{1}{x} f(e)=-3

to advance in the circuit, locate f(e^{4}).

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:

\(0\)