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

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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**

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.

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.**