Given the expression e^x + x^3 - x^2 + x prove that the derivative of te expression is positive.
f(x) = e^x + x^3 -x^2 + x
f'(x) = e^x + 3x^2 -2x + 1
Let us add and subtract x^2
==> f'(x) = e^x + 3x^2 + x^2 -2x +1 - x^2
==> f'(x) = e^x + 2x^2 + (x-1)^2
Then f'(x) > 0
Derivative of (e^x + x^3 - x^2 + x) is
(e^x + x^3 - x^2 + x)' = e^x+(3x^2-2x+1) .
= e^x+2x^2 +(x-1)^2. Obviously each of these 3 terms are positive.
Hence the derivative of (e^x+x^3-x^2+x) is always posiive.
We'll associate a function to the given expression:
f(x) = e^x + x^3 - x^2 + x
f'(x)=(e^x + x^3 - x^2 + x)'
We can combine the last 3 terms:
If we'll add (1),(2),(3):
So, f'(x) it's obviously positive.
The derivative of the expression e^x+x^3-x^2+x is e^x+3x^2-2x+1
Now e^x+3x^2-2x+1= e^x+2x^2+x^2-2x+1= e^x+2x^2+(x-1)^2
e^x is positive whether x is positive of negative. x^2 also is always positive irrespective of whether x is positive or negative. (x-1)^2 is also never negative irrespective of x being negative or positive.
Therefore the expression is the sum of three positive terms and therefore positive.