Let f(x) = 3+4x-x^{3}. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).
1. f is increasing on the intervals?
2. f is decreasing on the intervals?
f is increasing where f' is positive
f' = 4-3x^2
So, where is 4-3x^2 > 0?
similarly for decreasing.
max/min where 4-3x^2 = 0
To determine the intervals on which f(x) is increasing or decreasing, we need to find the first derivative of f(x) and analyze its sign.
1. To find where f(x) is increasing, we look for intervals where the first derivative is positive (greater than zero).
First, find the first derivative of f(x) by differentiating f(x) with respect to x:
f'(x) = d/dx (3 + 4x - x^3)
= 0 + 4 - 3x^2
= 4 - 3x^2
Set f'(x) > 0 to determine the intervals where f(x) is increasing:
4 - 3x^2 > 0
3x^2 < 4
x^2 < 4/3
|x| < √(4/3)
Since the inequality involves an absolute value, we consider both the positive and negative square roots:
x < √(4/3) and x > -√(4/3).
Therefore, f(x) is increasing on the intervals (-∞, -√(4/3)) and (√(4/3), +∞).
2. To find where f(x) is decreasing, we look for intervals where the first derivative is negative (less than zero).
Set f'(x) < 0 to determine the intervals where f(x) is decreasing:
4 - 3x^2 < 0
3x^2 > 4
x^2 > 4/3
|x| > √(4/3)
Again, considering both the positive and negative square roots:
x > √(4/3) and x < -√(4/3).
Therefore, f(x) is decreasing on the interval (-√(4/3), √(4/3)).
To find the x-coordinates of all relative maxima and minima, we need to find the critical points of f(x), which are the points where f'(x) = 0 or is undefined. In this case, f'(x) is defined for all x, so it becomes zero at the relative extrema.
Set f'(x) = 0 to find the critical points:
4 - 3x^2 = 0
3x^2 = 4
x^2 = 4/3
x = ± √(4/3)
Therefore, the x-coordinates of the relative maxima and minima are ± √(4/3).