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

Asked on by doorsreb

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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