# \n$f(x)=\\frac{1}{x}+1$ $f(1)=4$\nto advance in the circuit, locate $f(e)$.
# \n$f(x)=\\frac{1}{x}+1$ $f(1)=4$\nto advance in the circuit, locate $f(e)$.
Answer
Explanation:
Step1: Integrate (f^{\prime}(x))
$$ \begin{align*} f(x)&=\int\left(\frac{1}{x}+1\right)dx\ &=\int\frac{1}{x}dx+\int 1dx\ &=\ln|x|+x + C \end{align*} $$
Step2: Use the initial condition (f(1) = 4)
Substitute (x = 1) and (f(1)=4) into (f(x)=\ln|x|+x + C) $$ \begin{align*} 4&=\ln(1)+1 + C\ 4&=0 + 1+C\ C&=3 \end{align*} $$ So (f(x)=\ln x+x + 3) (since (x>0) as (\ln x) is considered here).
Step3: Find (f(e))
Substitute (x = e) into (f(x)=\ln x+x + 3) $$ \begin{align*} f(e)&=\ln(e)+e + 3\ &=1+e + 3\ &=e + 4 \end{align*} $$
Answer:
(e + 4)