Calculus: Early Transcendentals 8th Edition

$f(\frac{\sqrt 3}{3})\approx0.57$ is the absolute maximum value. $f(0)=0$ is the absolute minimum value.
The Closed Interval Method 1) Function $f$ must be continuous on the closed interval $[a, b]$ 2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints. 3) Compare the values. - The largest is the absolute maximum value. - The smallest is the absolute minimum value. $$f(t)=\frac{\sqrt t}{1+t^2},\hspace{1.5cm}[0, 2]$$ First, $f(t)$ is continuous on the interval $[0,2]$, so the Closed Interval Method can be applied here. 1) Find critical points of $f$ $$f'(t)=\frac{\frac{1}{2\sqrt t}(1+t^2)-\sqrt t\times2t}{(1+t^2)^2}$$ $$f'(t)=\frac{\frac{1+t^2-2t\sqrt t\times2\sqrt t}{2\sqrt t}}{(1+t^2)^2}$$ $$f'(t)=\frac{1+t^2-4t^2}{2\sqrt t(1+t^2)^2}$$ $$f'(t)=\frac{1-3t^2}{2\sqrt t(1+t^2)^2}$$ We can easily see that $f'(t)$ does not exist when $2\sqrt t(1+t^2)^2=0$, which means $t=0$ or $1+t^2=0$. However, since $1+t^2\gt0$ for all $t$, only $t=0$ is a critical point. Yet, $t=0$ does not lie in the interval $(0,2)$ given, so it will not be counted here. Also, $$f'(t)=0$$ $$1-3t^2=0$$ $$t^2=\frac{1}{3}$$ $$t=\pm\frac{\sqrt 3}{3}$$ These values also are critical points. However, only $t=\frac{\sqrt 3}{3}$ lies in the interval $(0,2)$, so only it will be evaluated next. 2) Calculate the values of $f$ at critical points and endpoints. - Critical points: $t=\frac{\sqrt 3}{3}$ $$f(\frac{\sqrt 3}{3})=\frac{\sqrt{\frac{\sqrt 3}{3}}}{1+(\frac{\sqrt 3}{3})^2}\approx0.57$$ - End points: $t=0$ and $t=2$ $$f(0)=\frac{\sqrt 0}{1+0^2}=0$$ $$f(2)=\frac{\sqrt 2}{1+2^2}=\frac{\sqrt 2}{5}\approx0.283$$ 3) Comparison Among the results: - $f(\frac{\sqrt 3}{3})\approx0.57$ is the largest, so $f(\frac{\sqrt 3}{3})\approx0.57$ is the absolute maximum value. - $f(0)=0$ is the smallest, so$f(0)=0$ is the absolute minimum value.