Question

Find all relative extrema and points of inflection for the following function...
h(X)= X^2+5X+4/ X-1
min=
max=
inflection points=

Answer 1

I will assume you meant:

h(x) = x^2 + 5x + 4/(x-1)

h ' (x) = 2x + 5 - 4/(x-1)^2
= 0 for max/min
2x + 5 = 4/(x-1)^2
(2x+5)(x^2 - 2x + 1) = 4
2x^3 - 4x^2 + 2x + 5x^2 - 10x + 5 = 4
2x^3 + x^2 - 8x + 1 = 0
hard to solve, Wolfram has this
http://www.wolframalpha.com/input/?i=2x%5E3+%2B+x%5E2+-+8x+%2B+1+%3D+0

if you meant it the way you typed it ...
h '(x) = 2x + 5 - 4/x^2 - 0
= 0

2x^3 + 5x^2 - 4 = 0
an equally messy cubic ....
http://www.wolframalpha.com/input/?i=2x%5E3+%2B+5x%5E2+-+4+%3D+0

are you expected to be able to solve a cubic ?