Question 760261
Q:
How do i find the range of f(x)={{{1/(x^2+3x+5)}}}
-----------------------------------------------------------------
A:
Let y = {{{1/(x^2+3x+5)}}}
Solve for x in terms of y.
{{{x^2 + 3x + 5}}} = {{{1/y}}}
{{{x^2 + 3x + (5 - 1/y) = 0}}}
Use the Quadratic Formula to solve for x.
{{{x = (-3 +- sqrt( 3^2-4*1*(5 - 1/y) ))/(2*1) }}}
{{{x = (-3 +- sqrt( -11 + 4/y ))/2 }}}
The expression inside the square root is {{{-11 + 4/y}}}.
{{{-11 + 4/y}}} ≥ 0
{{{(4 - 11y)/y}}} ≥ 0
Using Sign Test
0 < y &#8804; {{{4/11}}}
In interval notation, the range is (0, 4/11].