Find the intervals on which f(x) is increasing, decreasing and local extrema for f(x)=x^3-3x+7
To find the intervals on which f(x) is increasing or decreasing, we need to find the critical points of the function f(x). We do this by finding the derivative of f(x).
f'(x) = 3x^2 - 3
To find the critical points, we set f'(x) = 0 and solve for x:
3x^2 - 3 = 0
x^2 - 1 = 0
(x - 1)(x + 1) = 0
So the critical points are x = 1 and x = -1.
Since the derivative f'(x) = 3x^2 - 3 is a concave up parabola (with a positive leading coefficient), the function is increasing to the right of the critical points and decreasing to the left of them.
Therefore, the intervals on which f(x) is increasing are (-∞, -1) and (1, ∞).
The intervals on which f(x) is decreasing are (-1, 1).
To find the local extrema, we can evaluate f(x) at the critical points.
For x = -1:
f(-1) = (-1)^3 - 3(-1) + 7 = -1 + 3 + 7 = 9
For x = 1:
f(1) = (1)^3 - 3(1) + 7 = 1 - 3 + 7 = 5
So there is a local minimum at (-1, 9) and a local maximum at (1, 5).