Given the function g(x)= (3x) / (2+5x^2 ) find its region of increase and decrease. Give open intervals.
To find the region of increase and decrease for a given function, we need to find its derivative and determine where it is positive or negative.
First, let's find the derivative of the function g(x). Using the quotient rule, the derivative of g(x) can be calculated as follows:
g'(x) = [(2+5x^2)(3) - (3x)(10x)] / (2+5x^2 )^2
Simplifying further:
g'(x) = (6 + 15x^2 - 30x^2) / (2+5x^2 )^2
= (6 - 15x^2) / (2+5x^2 )^2
Now, to determine the regions of increase and decrease, we need to find where the derivative g'(x) is positive or negative.
For the derivative g'(x) to be positive, we need the numerator, 6 - 15x^2, to be greater than zero. Solving the inequality:
6 - 15x^2 > 0
15x^2 < 6
x^2 < 6/15
x^2 < 2/5
Taking the square root on both sides, we get:
|x| < √(2/5)
Now, to find the region of increase (where g(x) is increasing), we look for values of x such that g'(x) > 0. From the inequality x^2 < 2/5, we can determine the open intervals where g(x) is increasing:
-√(2/5) < x < √(2/5)
Similarly, for the derivative g'(x) to be negative, we need the numerator, 6 - 15x^2, to be less than zero. Solving the inequality:
6 - 15x^2 < 0
15x^2 > 6
x^2 > 6/15
x^2 > 2/5
Taking the square root on both sides, we get:
|x| > √(2/5)
Again, to find the region of decrease (where g(x) is decreasing), we look for values of x such that g'(x) < 0. From the inequality x^2 > 2/5, we can determine the open intervals where g(x) is decreasing:
x < -√(2/5) or x > √(2/5)
Therefore, the region of increase is (-√(2/5), √(2/5)), and the region of decrease is (negative infinity, -√(2/5)) and (√(2/5), positive infinity).