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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

  • 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 $f(2)=\large\frac{2}{3}$
answered Jul 31, 2013 by meena.p

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