What are the relative minimums, maximums, and points of inflection?

Question

f(x) = x^3 + x + 4/x

Answer 1

Take the derivative and set to zero.

f(x) = x^3 + x + 4/x
f'(x) = 3x² + 1 - 4/x²

0 = 3x² + 1 - 4/x²
Multiply both sides by x²
0 = 3x^4 + x² - 4

Now you can set k = x²
3k² + k - 4 = 0

(3k + 4)(k - 1) = 0
k = 1
k = -4/3

x = √1 = ±1
x = √(-4/3) = ±2i/√3

So the only real answers are x = -1 or x = 1

The y values would be:
f(-1) = (-1)^3 + (-1) + 4/(-1)
f(-1) = -1 -1 - 4 = -6

f(1) = (1)^3 + 1 + 4/1
f(1) = 1 + 1+ 4 = 6

Points (-1, -6), (1, 6)

Answer 2

f'(x) = 3x^2+1-4/x^2

max and min are where f'(x)=0

3x^2+1-4/x^2=0
multiply by x^2
3^x4+x^2-4=0
(3x^2+4)(x^2-1)=0

only real solutions are x=+/-1
complex ones are x=+/-2i/√3

f''(x)= 6x+8/x^3
at(1,6) f''(x) is +tive so minimum
at(-1,-6) f''(x) -tive so maximum

points of inflection where f''(x)=0
hence 6x+8/x^3=0
gives no real solutions