Demonstrate that f(x)= e^x + x^3 - x^2 + x is an increasing function.
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.