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

If the first derivative of the function is positive, then the function is increasing.

Let's calculate f'(x):

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

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

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

We'll re-write the terms of the expression of the first derivative, to complete the squares:

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

We'll combine the last 3 terms, because we've noticed that they are the result of squaring the binomial (x-1).

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

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

**Since each term of the expression of f'(x) is positive, the sum of positive terms is also a positive expression. The expression of f'(x) it's obviously>0, so f(x) is an increasing function.**

To prove y = e^x +x^3-x^2+x is increasing function.

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

We know that e^x is always incresing for all x as (e^x)' >0 for all x.

3x^2-2x+1 has the discriminant (-2)^2 - 3*3*1 = 4-9 = -5 <0. Therefore 3x^2-2x+1 > 0 for allx.

Therefore from (1) y ' = e^x +(3x^2-2x+1) > 0 for all x.

Therefore if y' > 0 , and y is a continuous function, then y is an increasing function.

Therefore e^x+x^3-x^2+x is an increasing function.