# Find the absolute maximum and absolute minimum values of $f$ on the given interval: $\;f(x)=\large\frac{x}{x+1},$$[1 , 2 ] Note: This is part 5th of a 7 part question, split as 7 separate questions here. ## 1 Answer Toolbox: • To find the absolute maximum and minimum values of a continuous function f on a closed interval [a,b] • (i) Find the values of the critical numbers of f in (a,b) • (ii) Find the value of f(a) and f(b) • (iii) The largest of the values from (i) and (ii) is the absolute maximun value, the smallest of these values is the absolute minimum value. f(x)=\large\frac{x}{x+1} is continous in (-1,2] Step 1: It is not defined at x=-1\; (\to \infty) Step 2: f'(x)= \large\frac{(x+1)(1)-x(1)}{(x+1)^2} \qquad=\large\frac{1}{(x+1)^2} \neq$$0$ for $x \in (-1,2]$
There are no critical points in the interval. The values of f at the endpoints are $f(-1)=\infty,f(2)=\large\frac{2}{3}$
There is no absolute maximum. The absolute minimum is $f(2)=\large\frac{2}{3}$