# Verify the monotony of the function y=x^3-x^2+x+e^x?

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In order to verify the monotony of y = f(x), we'll have to demonstrate that the first derivative of the function is positive or negative.

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

We'll re-write f'(x):

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

We can group the last 3 terms, because we've noticed that they are the development of the binomial square .

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

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

We'll re-write f'(x):

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

Since (x-1)^2>0 and e^x+2x^2 > 0, then e^x+2x^2+(x-1)^2>0

The expression of f'(x) is positive, for any value of x, so f(x) is an increasing function.

y = e^x+x^3-x^2+x

To discuss the monotony of the function, we first find dy/dx and prove that it is positive. So the function y is an creasing funtion for all x.

y = e^x+x^3-x^2+x.

dy/dx = (e^x+x^3-x^2+x)'

dy/dx = e^x+3x^2-2x+1.

dy/dx = e^x+3(x^2-(2/3)x)+2

dy/dx = e^x +3 (x-1/3)^2- 3(1/3)^2+1

dy/dx= e^x +3(x-1/3)^2 + 2/3.

Now consider the right side: The first term e^x > 1 for all x. The second term 3(x-1/3)^2 > = 0 for all x And the 3rd term 2/3 is a constant positivr term

Therefore dy/dx = e^x+x^3-x^2+x > 0 for all x.

Therefore y = e^x +x^3 -x^2+x is monotonously increasing function for all x.