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

3 Answers | Add Yours

justaguide's profile pic

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's profile pic

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's profile pic

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.

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question