let f be the function defined by f(x)=12x^2/3 -4X
a)find the intervals on which f is increasing
I got the rest of the question but I'm still not getting the increasing/decreasing thing
Well, let's break it down nice and simple. The function f(x) is defined as f(x) = 12x^(2/3) - 4x.
To determine where the function is increasing, we need to find the intervals where the derivative is positive.
First, let's take the derivative of f(x).
f'(x) = (2/3)(12)(x)^(2/3 - 1) - 4
Simplifying, we have:
f'(x) = 8(x)^(-1/3) - 4
Now, to find where f'(x) > 0, we can set it equal to 0 and solve for x:
8(x)^(-1/3) - 4 > 0
Dividing both sides by 8, we get:
(x)^(-1/3) > 1/2
To get rid of that pesky negative exponent, we can raise both sides of the inequality to the power of -3:
1/(x) > (1/2)^(-3)
1/x > 8
Now, we can multiply each side of the inequality by x to isolate x:
1 > 8x
x < 1/8
So, the interval where f(x) is increasing is (-∞, 1/8).
I hope that explanation wasn't too derivative for you!
To determine where a function is increasing or decreasing, we can look at its derivative. In this case, let's find the derivative of the given function f(x):
f(x) = 12x^(2/3) - 4x
To find f'(x) (the derivative of f(x)), we apply the power rule and differentiate each term:
f'(x) = d/dx (12x^(2/3)) - d/dx (4x)
Using the power rule, the derivative of x^n is nx^(n-1), so:
f'(x) = (2/3)(12)x^((2/3)-1) - 4
Simplifying further gives:
f'(x) = 8x^(-1/3) - 4
Now, to find where f'(x) > 0 (positive), we solve the inequality:
8x^(-1/3) - 4 > 0
Adding 4 to both sides:
8x^(-1/3) > 4
Dividing both sides by 8:
x^(-1/3) > 1/2
Taking the cube of both sides (which preserves the inequality since 1/2 is positive):
x > (1/2)^3 = 1/8
Thus, the function f(x) is increasing for x > 1/8.
To determine the intervals on which a function is increasing or decreasing, you need to analyze the sign of its derivative. Recall that the derivative of a function represents its rate of change.
To find the derivative of the function f(x) = 12x^(2/3) - 4x, we apply the power rule and the constant rule of derivatives.
1. Apply the power rule:
d/dx (x^n) = n * x^(n-1)
For the first term, 12x^(2/3), the derivative is:
d/dx (12x^(2/3)) = 12 * (2/3) * x^((2/3) - 1) = 8x^(-1/3).
For the second term, -4x, the derivative is simply -4.
2. Combine the derivatives:
f'(x) = 8x^(-1/3) - 4.
Now, to find the intervals where f(x) is increasing, we need to analyze the sign of f'(x) in different regions.
3. Set f'(x) = 0:
8x^(-1/3) - 4 = 0.
To solve this equation, add 4 to both sides and then cube both sides:
8x^(-1/3) = 4
x^(-1/3) = 1/2
(1/x)^(1/3) = 1/2
1/x = (1/2)^3
1/x = 1/8
x = 8.
So, f'(x) = 0 at x = 8.
4. Choose test points:
Now, to determine the intervals on which f(x) is increasing or decreasing, we need to choose test points within these intervals.
Select a test point less than 8, say x = 0. Plugging this value into f'(x), we get:
f'(0) = 8(0)^(-1/3) - 4 = -4.
Select a test point greater than 8, say x = 10. Plugging this value into f'(x), we get:
f'(10) = 8(10)^(-1/3) - 4 = 1.14.
5. Analyze the intervals:
Based on the test points, we see that for x < 8, f'(x) < 0, and for x > 8, f'(x) > 0.
Therefore, f(x) is decreasing for x < 8 and increasing for x > 8.
To summarize, the function f(x) = 12x^(2/3) - 4x is increasing on the interval (8, ∞) and decreasing on the interval (-∞, 8).
As we saw in a previous posting, f(x) is increasing where f'(x) is positive.
A positive slope means that when x increases, f(x) increases. A negative slope means that as x increases, f(x) decreases.
f'(x) = 8x^(-1/3) - 4
So, we want 8/x^1/3 - 4 > 0
8/x^1/3 > 4
1/x^1/3 > 1/2
x^1/3 > 2
x > 8
Oops. Taking reciprocals reverses the direction of the inequality:
8/x^1/3 - 4 > 0
8/x^1/3 > 4
1/x^1/3 > 1/2
x^1/3 < 2
x < 8