# Demonstrate that the function f(x)=e^2x+x^3-2x^2+4x is monotonically increasing.

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### 2 Answers

A monotonically increasing function has a derivative that is always positive.

For f(x) = e^2x + x^3 - 2x^2 + 4x

f'(x) = 2*e^2x + 3x^2 - 4x + 4

2*e^2x is always positive.

Let g(x) = 3x^2 - 4x + 4

g'(x) = 6x - 4

6x - 4 = 0

=> x = 2/3

g''(x) = 6 which is positive, this shows that g(x) has a minimum value at x = 2/3

g(2/3) = 3*(2/3)^2 - 4(2/3) + 4

=> 3*4/9 - 8/3 + 4

=> (12 - 24 + 36)/9

=> 24/9 which is positive.

This gives us that 3x^2 - 4x + 4 is always positive

As both for f'(x) = 2*e^2x + 3x^2 - 4x + 4 both 2*e^2x and 3x^2 - 4x + 4 are always positive, f'(x) is always positive.

**Therefore f(x)=e^2x+x^3-2x^2+4x is monotonically increasing.**

Since we have to use the first derivative to verify if the function is monotonic, first we need to check if the function is continuous. Since the given function is an algebraic sum of elementary continuous functions, then f(x) is continuous.

In order to decide the monotony of f(x), we'll have to demonstrate that the first derivative does not change its sign.

We'll compute f'(x)=2e^2x+3x^2-4x+4

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

f'(x)=2e^2x+2x^2+x^2-4x+4

We notice that the sum of the last 3 terms represents a perfect square:

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

x^2-4x+4 = (x-2)^2

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

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

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

**We notice that the expression of f'(x) keeps it's positive sign, for any value of x, therefore f(x) is a monotonically increasing function.**