Find the intervals on which f(x) is increasing, decreasing and local extrema for f(x)=x^3+5x+7
To find the intervals on which f(x) is increasing and decreasing, we first need to find the derivative of f(x) and determine where the derivative is positive or negative.
Taking the derivative of f(x) = x^3 + 5x + 7 gives us:
f'(x) = 3x^2 + 5
To find where f'(x) = 3x^2 + 5 is positive or negative, we set it equal to 0 and solve for x:
3x^2 + 5 = 0
3x^2 = -5
x^2 = -5/3
Since x^2 cannot be negative, there are no real solutions to this equation. Therefore, f'(x) is either always positive or always negative.
Let's check the values of f'(x) at a few points to determine if it is positive or negative:
-1: f'(-1) = 3(-1)^2 + 5 = 8 (positive)
0: f'(0) = 3(0)^2 + 5 = 5 (positive)
1: f'(1) = 3(1)^2 + 5 = 8 (positive)
From the above calculations, we see that f'(x) is always positive. Therefore, f(x) is increasing for all x.
Next, let's find the local extrema. Since f(x) is continuously increasing, it does not have any local extrema.