University Calculus: Early Transcendentals (3rd Edition)

$f(x)$ has the absolute maximum value of $1.636$ at $x=4$ and the absolute minimum value of $1$ at $x=1$.
$$f(x)=\frac{1}{x}+\ln x\hspace{1cm}0.5\le x\le4$$ 1) Find the critical points of $f(x)$ Find $f'(x)$: $$f'(x)=\frac{(1)'x-1(x)'}{x^2}+\frac{1}{x}$$ $$f'(x)=\frac{0\times x-1\times 1}{x^2}+\frac{1}{x}$$ $$f'(x)=-\frac{1}{x^2}+\frac{1}{x}=\frac{x-1}{x^2}$$ $f(x)=0$ when $x-1=0$, or when $x=1$. $f(x)$ is undefined when $x^2=0$, or when $x=0$. Both $x=0$ and $x=1$ are critical points of $f(x)$. However, since $0\notin[0.5,4]$, we only take into consideration the critical point $x=1$ here. 2) Evaluate $f(x)$ at the critical points and endpoints: - For $x=0.5$: $$f(x)=\frac{1}{0.5}+\ln0.5=2+\ln0.5\approx1.307$$ - For $x=1$: $$f(x)=\frac{1}{1}+\ln1=1+0=1$$ - For $x=4$: $$f(x)=\frac{1}{4}+\ln4=0.25+\ln4\approx1.636$$ From the values of $f(x)$ above, we conclude that $f(x)$ has the absolute maximum value of $1.636$ at $x=4$ and the absolute minimum value of $1$ at $x=1$.