If f(x)= 3x+6, prove that f( f(x)) is an increasing function.

Asked on by cesther86

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To establish that a function is increasing, we'll have to do the first derivative test. To do the first derivative test, we'll have to determine the result of the composition of f(x) with f(x).

f(x)*f(x) = f(f(x))

We'll substitute in the expression of f(x), the variable x by the expression of f(x).

f(f(x)) = 3f(x) + 6

f(f(x)) = 3(3x+6) + 6

We'll remove the brackets:

f(f(x)) = 9x + 18 + 6

We'll combine like terms:

f(f(x)) = 9x + 24

Since we know the expression of f(f(x)), we can do the first derivative test.

f'(f(x)) = (9x + 24)'

f'(f(x)) = 9

If the first derivative is positive, then the original function is increasing.

Since the result of the first derivative test is positive, then f(f(x)) is an increasing function.

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

In the problem we are given the function f(x) = 3x+6.

Now to determine f (f(x)) we first write the inner function as 3x+6

f (f(x))

=> f (3x+6)

=> 3*(3x+6) +6

=> 9x+18 +6

=> 9x +24

Now to establish that this is an increasing function, we need to calculate the first derivative of f (f(x)).

This is the derivative of 9x+24 with respect to x. The result is the constant number 9 which is a positive number.

Therefore as the derivative is always positive we can say that the function is always increasing.

Therefore f (f(x)) is an increasing function.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

If f(x) = 3x+6.

So f(f(x)) = 3f(x)+6.

f(fx)) = 3(3x+6)+6

f(f(x)) = 9x+18+6

f(f(x)) = 9x+24

f(f(x)) = y = 9x+24.

If y is an increasing function, then dy/dx should be positive.

Difetiating both sides of y = 9x+24, we get:

dy/dx = (9x+24) = 9 which is positive. So f(f(x)) = y = 9x+24 is an increasing function of x.


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