# Verify if the function is increasing y = 3x^3 + 3^x .

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### 5 Answers

We have the function y = 3x^3 + 3^x . We have to verify that the function is increasing.

For an increasing function the first derivative is always positive.

Now y = 3x^3 + 3^x

y' = 9x^2 + 3^x * log 3

Now 9 x^2 is always positive, log 3 is always positive and 3^x is always positive.

Therefore y' = 9x^2 + 3^x * log 3 is always positive.

**This proves that y = 3x^3 + 3^x is an increasing function.**

Given the function y=3x^3 + 3^x.

We need to determine whether **y** is an increasing function.

To verify, we need to determine the first derivative ( y').

If y' is positive, then the function is increasing.

If y' is negative, then the function is decreasing.

Now, let us determine the first derivative:

y= 3x^3 + 3^x

==> y' = ( 3x^3)' + (3^x) '

= 9x^2 + (3^x)* ln 3

Now let us analyze y'.

We know that x^2 is always positive.

Then, 9x^2 also positive.

Also, 3^x is always positive.

Then, (3^x)* ln3 is positive.

Then, we conclude that y' is positive.

**Then, the function y is an increasing function for all R numbers.**

To verify if 3x^3+3^x is increasing.

We take the first derivative of the function and examine whether it is positive.

f(x) = 3x^3+3^x.

f'(x) = (3x^3 +3^x)'.

f'(x) = 3*3x^2+log3)3^x.

f'(x) = 3x^2 + (log3)3^x

Onthe right side the first term x^2 > = 0 for all x. Therefore 9*x^2 > = 0 for all x . 9x^2 = 0 at x= 0.

The second term (log3)3^x > 0 forall x as, 3^x > 0 for all x. log3 is also > 0 , as 3 > 1, log 3 > log 1 = 0.

Therefore f'(x) = 3x^2 +(3log3) 3^x > 0.

Therefore f(x) = 3x^3+3^x is an increasing function.

the proof of these kinds of questions is usually to make two assumption, a and b(a is greater than b), for x.

substitute a and b in to the equation then use the equation of b to subtract the equation of a.

if the answer is positive then its increasing, else its decreasing.

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

Let's calculate f'(x) = (3x^3 + 3^x)':

f'(x) = 9x^2 + 3^x*ln3

Since 9x^2>0

and 3^x*ln3 > 0

The sum of 2 positive amounts is also positive:

9x^2 + 3^x*ln3 > 0

f'(x)>0

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