The function f given by f(x) = x^3 + 12x – 24 is?
Increasing for all x's? Cause that's what I think..
f ' (x) = 3x^2 + 12
which is clearly positive for any value of x you give me.
So, yes, you are correct
Well, the function f(x) = x^3 + 12x - 24 might not be increasing for all x's. But don't worry, if it's feeling down, we can always try to boost its confidence with a motivational speech!
To determine whether the function f(x) = x^3 + 12x - 24 is increasing for all x, we need to find the derivative of the function and check its sign.
The derivative of the function f(x) = x^3 + 12x - 24 can be found by taking the derivative of each term separately using the power rule.
f'(x) = 3x^2 + 12
To determine the sign of the derivative, we can check the sign of each term. Since the coefficient in front of x^2 is positive, 3x^2 is positive for all x. Similarly, the constant term 12 is also positive. Therefore, f'(x) = 3x^2 + 12 is positive for all x.
When the derivative is positive for all x, it means the function is increasing for all x. Hence, you are correct in thinking that the function f(x) = x^3 + 12x - 24 is increasing for all x.
To determine whether the function f(x) = x^3 + 12x - 24 is increasing for all x, you need to analyze the derivative of the function.
To find the derivative of f(x), you can apply the power rule for differentiation, which states that if y = ax^n, then dy/dx = nax^(n-1).
Applying the power rule to the function f(x), we get:
f'(x) = 3x^2 + 12
Now, to determine whether f(x) is increasing for all x, you need to check the sign of the derivative, f'(x), for all x.
In this case, the derivative f'(x) = 3x^2 + 12 is a quadratic polynomial. To find the critical points where the derivative changes sign, you can set f'(x) = 0 and solve for x:
3x^2 + 12 = 0
x^2 + 4 = 0
(x + 2)(x - 2) = 0
From this equation, we find two critical points: x = -2 and x = 2.
Now, you can create a sign chart to analyze the sign of the derivative f'(x) for different intervals:
Interval (-∞, -2): Pick a test point, e.g., x = -3
f'(-3) = 3(-3)^2 + 12 = 27 > 0
Interval (-2, 2): Pick a test point, e.g., x = 0
f'(0) = 3(0)^2 + 12 = 12 > 0
Interval (2, +∞): Pick a test point, e.g., x = 3
f'(3) = 3(3)^2 + 12 = 39 > 0
Since the sign of f'(x) is positive for all intervals, we can conclude that the function f(x) = x^3 + 12x - 24 is indeed increasing for all x.
Note: It's always a good idea to double-check your calculations and consider the behavior of the function at critical points and at infinity.