0
$\begingroup$

Find absolute extrema of the function $ \ f(x)=1-|x-1| , \ \ x \in [-9,4] \ $ on the closed interval.

Answer:

$f(x)= \begin{array} (2-x) , \ \ if \ x \in [1,4] \\ \ \ x , \ \ if \ \ x \in [-9,1] \end{array} \ $

The function has not critical point.

Thus the extrema occurs at the end points $ \ x=1, x=4, x=-9 \ $

Now,

$ f(1)=1, \ f(-9)=-9 , \ f(4)=-2 \ $

Thus,

absolute minima $ \ f(-9)=-9 \ $

absolute maxima $ \ f(1)=1 \ $

I need confirmation of my work.

$\endgroup$
2
  • 1
    $\begingroup$ This seems fine to me. You might add, that since $f$ is continuous and $[-9,4]$ is a compact intervall (it is bounded and closed), your function $f$ maps onto a maximum and minimum. $\endgroup$
    – Cornman
    Jul 11, 2018 at 1:36
  • $\begingroup$ To confirm your calculations, you could draw the graph of the function. $\endgroup$
    – MasB
    Jul 11, 2018 at 1:45

1 Answer 1

1
$\begingroup$

Your work is correct. The absolute maximum value of $1$ is attained at $x=1$ and the absolute minimum value of $-9$ is attained at $x=-9.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .