What is the minimum value of #g(x) = (x-1)/(x^2+4)?# on the interval #[-2,2]#?

1 Answer
Dec 22, 2016

Minimum value is at #x=1-sqrt 5 approx "-"1.236#;

#g(1 - sqrt 5) = -(1+ sqrt 5)/(8) approx "-"0.405#.

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 #g(x)# at any #x in ["-2",2]# that makes #g'(x)=0#, as well as at #x="-2"# and #x=2#.

First: what is #g'(x)#? Using the quotient rule, we get:

#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 #x#-values is in #["-2",2]#, and that is #x=1-sqrt 5#.

Now, we compute:

1. #g("-2") = ("-"2-1)/(("-2")^2+4)="-3"/8="-"0.375#

2. #g(1 - sqrt 5) = (1 - sqrt 5 -1)/((1 - sqrt 5)^2+4)=("-"sqrt 5)/(1-2 sqrt 5 + 5+4)#
#color(white)(g(1 - sqrt 5)) = -(sqrt 5)/(10-2sqrt 5) =-(sqrt 5)/((2)(5-sqrt5)) * color(blue)((5+sqrt 5)/(5+ sqrt 5))#
#color(white)(g(1 - sqrt 5)) =-(5+5 sqrt 5)/(2 * (25-5)#
#color(white)(g(1 - sqrt 5)) =-(5(1+sqrt5))/(40)=-(1+sqrt 5)/(8) approx "-"0.405#

3. #g(2)=(2-1)/(2^2+4)=1/8 =0.125#

Comparing these three values of #g(x)#, we see that #g(1-sqrt 5)# is the smallest. So #-(1+ sqrt 5)/8# is our minimum value for #g(x)# on #["-"2, 2]#.