# Given the function f(x)=x^3-3x+3arctan x, prove that f increases if x is real?

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### 2 Answers

The function given is f(x) = x^3 - 3x + 3* arc tan x. If a function is always increasing its first derivative should always be positive. The first derivative if f(x)

f'(x) = [x^3 - 3x + 3* arc tan x]'

=> 3x^2 - 3 + 3/(1 + x^2)

=> 3*(x^2 - 1 + 1/(1 + x^2))

=> 3*(x^2 - 1)(x^2 + 1) + 1)/(x^2 + 1)

=> 3*(x^4 - 1+ 1)/(x^2 + 1)

=> 3*x^4 /(x^2 + 1)

Now, if x is real x^2 and x^4 are positive. In 3*x^4/(1 + x^2) all the terms are positive for real values of x. This verifies that f'(x) is positive for all real values of x.

**As the first derivative of f(x) is positive for all real values of x, f(x) is an increasing function.**

To prove that f is strictly increasing, we'll have to check if the 1st derivative is positive for any real value of x.

We'll determine the 1st derivative of f(x):

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

f'(x) = (3x^2 + 3x^4 - 3 - 3x^2 + 3)/(1+x^2)

Combining and eliminating like terms inside bracktes, we'll have:

f'(x) = 3x^4/(1+x^2)

We notice that both numerator and denominator are positive for any value of x.

f'(x) = 3x^4/(1+x^2) > 0

**If the 1st derivative of f(x) is strictly positive, then the function f(x) = x^3-3x+3arctan x is strictly increasing for any real value of x.**