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
and 3^x*ln3 > 0
The sum of 2 positive amounts is also positive:
9x^2 + 3^x*ln3 > 0
The expression of f'(x) is positive, for any value of x, so f(x) is an increasing function.