f(x) = e^x + x^3 - x^2 +x

The function f(x) is increasing if the first derivative f'(x) >0

Let us dtermine f'(x):

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

= e^x + 2x^2 + x^2 -2x +1

= e^x + 2x^2 + (x-1)^2

All terms are positive values, then f'(x)>0, then f(x) is increasing.

f(x) = e^x+x^3-x^2+x.= (e^x+x)+(x^3-x^2)

f'(x) = e^x+(3x^2-2x+1) > 0 as the expression i bracket 3x^2-2x+1 is always positive since the discriminant of this quadratic 3x^2-2x+1 is (-2)^2-4*3*1 = 4-12 = -8 a negative quantity.

Therefore f(x) = e^x+x^3-x^2+x is always increasing.

In order to prove that f(x) is an increasing function, we have to demonstrate that the first derivative of the function is positive.

Let's calculate f'(x) = e^x + 3x^2 - 2x + 1 =

f'(x) = e^x + 2x^2 + x^2 - 2x + 1

We notice that the last 3 terms are the resut of expanding the square, using the formula:

(a + b)^2 = a^2 + 2ab + b^2

f'(x) = e^x + 2x^2 + (x - 1)^2 > 0

The last form for f'(x) it's obviously > 0, so f(x) is an increasing function.