F (x) = x^3+8x-19, show that the equation f (x) = 0 has one real root

1) Find whether the function, f (x) has a local or absolute maximum or minimum.

Calculate the first derivative: f ' (x) = 3x^2 + 8

Maximum or minimum = > f ' (x) = 0 = > 3x^2 + 8 = 0 = > 3x^2 = - 8

Which has not a real solution because x^2 is always positive.

Then there is not maximum or minimum.

Then that function is or increasing or decreasing in all the domain and it can have maximum one real root.

Yet we have to show if there is one root or none root.

if you make x = 1 you find f (x) = 1^3 + 8 (1) - 19 = 1 + 8 - 19 = - 10

If you make x = 2, you find f (x) = 2^3 + 8 (2) - 19 = 8 + 16 - 19 = 5

Then f (x), given that it is continuos, will have a zero between x = 1 and x = 2, which corresponds to the only root of the function.