Prove that the function f(x) = 10x^7 + 7^x is increasing.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

A function is increasing if the derivative of the function is positive.

Now for f(x) = 10x^7 + 7^x

f'(x) = 10*7*x^6 + (ln 7)*7^x.

=> f'(x) = 70x^6 + (ln 7)*7^x

Both the terms 70x^6 and (ln 7)*7^x can only take on positive values.

Therefore f'(x) is positive and we can say that f(x) = 10x^7 + 7^x is increasing

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To prove that the function f(x) = 10x^7 + 7^x is increasing.

A function f(x) is increasing  in any interval if if derivative  f'(x) is positive in  the interval.

10x^7 is is an increasing function as its derivative (10x^7)' = 10*6x^2 > 0 for all x.

Similarly 7^x is an increasing function as its derivative (7^x )' = (log7)7^x is also positive for all x.

Therefore 10x^7 + 7^x is an increasing function.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

In order to verify if the function f(x) is increasing, we'll have to demonstrate that the first derivative of the function is positive.

Let's calculate f'(x) = (10x^7 + 7^x)'

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

f'(x) = (10x^7)' + (7^x)'

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

f'(x) = 70x^6 + 7^x*ln7

Since 70x^6 >0 and 7^x*ln7 > 0 (7^x>0 and ln 7 = 1.94) , then 70x^6 + 7^x*ln7 > 0

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