What is the minimum value of #g(x) = (x-1)/(x^2+4)?# on the interval #[-2,2]#?
1 Answer
Minimum value is at
Explanation:
On a closed interval, the possible locations for a minimum will be:
- a local minimum inside the interval, or
- the endpoints of the interval.
We therefore compute and compare values for
First: what is
#g'(x)=((1)(x^2+4)-(x-1)(2x))/(x^2+4)^2#
#color(white)(g'(x))=(x^2+4-2x^2+2x)/(x^2+4)^2#
#color(white)(g'(x))=-(x^2-2x-4)/(x^2+4)^2#
This will equal zero when the numerator is zero. By the quadratic formula, we get
#x^2-2x-4=0" "=>" "x=1+-sqrt 5 approx {"-1.236", 3.236}#
Only one of these
Now, we compute:
1.
2.
3.
Comparing these three values of