Let F be the function defined by F(X)=12X to the 2/3 power minus 4X Find the intervals of which F is increasing? Find X & Y coordinates of all relative max. pts? Find X and Y Coordinates of all relative min pts? Find the intervals of which F is concave down? Sketch a graph of F
When the second derivative is negative
To find the intervals on which the function F is increasing, we need to determine where the first derivative of the function is positive.
1. Find the derivative of F(x):
F'(x) = (2/3)*(12X)^(2/3-1) * 12X^(1/3) - 4
Simplifying this expression, we get:
F'(x) = (8/3) * X^(1/3) - 4
To find the intervals where F is increasing, we need to solve the inequality F'(x) > 0.
2. Solve the inequality:
(8/3) * X^(1/3) - 4 > 0
Adding 4 to both sides:
(8/3) * X^(1/3) > 4
Multiplying both sides by 3/8:
X^(1/3) > 3/2
Taking the cube of both sides to remove the exponent:
X > (3/2)^3
X > 27/8
Therefore, the function F is increasing for all x-values greater than 27/8 or x > 27/8.
To find the relative maximum points, we need to determine where the first derivative changes from positive to negative.
3. Set F'(x) = 0 and solve for x:
(8/3) * X^(1/3) - 4 = 0
Adding 4 to both sides:
(8/3) * X^(1/3) = 4
Multiplying both sides by 3/8:
X^(1/3) = 3/2
Cubing both sides to remove the exponent:
X = (3/2)^3
X = 27/8
Now we need to find the corresponding y-coordinate by substituting this x-value into the original function:
F(27/8) = 12 * (27/8)^(2/3) - 4 * (27/8)
F(27/8) = 12 * (27/8)^(2/3) - 27/2
So the relative maximum point is (27/8, F(27/8)) = (27/8, 27/2).
To find the relative minimum points, we need to determine where the first derivative changes from negative to positive.
4. Set F'(x) = 0 and solve for x:
(8/3) * X^(1/3) - 4 = 0
Adding 4 to both sides:
(8/3) * X^(1/3) = 4
Multiplying both sides by 3/8:
X^(1/3) = 3/2
Cubing both sides to remove the exponent:
X = (3/2)^3
X = 27/8
Now we need to find the corresponding y-coordinate by substituting this x-value into the original function:
F(27/8) = 12 * (27/8)^(2/3) - 4 * (27/8)
F(27/8) = 12 * (27/8)^(2/3) - 27/2
So the relative maximum point is (27/8, F(27/8)) = (27/8, 27/2).
To find the intervals on which F is concave down, we need to determine where the second derivative of the function is negative.
5. Find the second derivative of F(x):
F''(x) = d/dx [(8/3) * X^(1/3) - 4]
F''(x) = (8/3) * d/dx [X^(1/3)] - 0
F''(x) = (8/3) * (1/3) * X^(-2/3)
F''(x) = (8/9) * X^(-2/3)
To determine where F is concave down, we need to solve the inequality F''(x) < 0.
6. Solve the inequality:
(8/9) * X^(-2/3) < 0
X^(-2/3) < 0
Since X^(-2/3) is always positive, there are no intervals on which F is concave down.
Now, let's sketch the graph of F to visualize the function.
To determine the intervals where the function F is increasing, we need to find the intervals where the first derivative of F is positive.
1. Calculate the first derivative of F:
To find the derivative of F(X), we need to apply the power rule and the constant rule:
F'(X) = (2/3) * 12X^(2/3 - 1) - 4
= (2/3) * 12X^(-1/3) - 4
= 8X^(-1/3) - 4
2. Set the first derivative equal to zero:
8X^(-1/3) - 4 = 0
3. Solve for X:
8X^(-1/3) = 4
X^(-1/3) = 4/8
X^(-1/3) = 1/2
(X^(-1/3))^3 = (1/2)^3
X^(-1) = 1/8
1/X = 1/8
X = 8
The critical point is X = 8. We have divided the number line into three intervals: (-∞, 8), (8, 8), and (8, +∞). If we choose a test value (less than 8) in the first interval or a test value (greater than 8) in the third interval, plug it into the first derivative, we can determine the sign of the derivative and thus identify the intervals of increase.
Let's test X = 7 (in the interval (-∞, 8)):
F'(7) = 8(7^(-1/3)) - 4 = 8(1/7) - 4 = 8/7 - 4 > 0
Since the derivative is positive, F is increasing in the interval (-∞, 8).
To find the X and Y coordinates of all relative maximum points, we look for local maximums by examining the critical points and the endpoints of the intervals.
1. Critical point:
X = 8
Substitute this value back into the original function F(X):
F(8) = 12(8)^(2/3) - 4(8) = 12(4) - 32 = 48 - 32 = 16
The relative maximum point is (8, 16).
2. Endpoints:
We also need to check the values at the endpoints of the intervals (-∞, 8) and (8, +∞).
For X = 0 (interval -∞, 8):
F(0) = 12(0)^(2/3) - 4(0) = 0 - 0 = 0
For X = ∞ (interval 8, +∞):
As X approaches infinity, the function F(X) also tends to infinity.
There are no other relative maximum points in the given function.
To find the X and Y coordinates of all relative minimum points, we again examine the critical points and the endpoints.
1. Critical point:
We already found a critical point at X = 8, which was a relative maximum. Therefore, there are no relative minimum points.
2. Endpoints:
For X = 0 (interval -∞, 8):
We already calculated F(0) = 0, but this is not a relative minimum point.
For X = ∞ (interval 8, +∞):
Again, as X approaches infinity, F(X) tends to infinity, so there are no relative minimum points.
To find the intervals where F is concave down, we need to determine where the second derivative is negative.
1. Calculate the second derivative of F:
To find the second derivative of F(X), we need to differentiate F'(X) with respect to X:
F''(X) = d/dX (8X^(-1/3) - 4)
= (-1/3) * 8X^(-1/3 - 1)
= (-1/3) * 8X^(-4/3)
= (-8/3) * X^(-4/3)
2. Determine the intervals:
Since the second derivative is negative when (-8/3) * X^(-4/3) < 0, we know that X^(-4/3) must be positive. This is because multiplying a negative constant with a positive value will yield a negative value.
X^(-4/3) > 0
Taking the reciprocal of both sides (which preserves the inequality sign when dealing with positives)
1/X^(4/3) > 0
We know that the function 1/X^a is positive when a > 0. Therefore, the inequality holds for all X ≠ 0.
Thus, F is concave down for the interval (-∞, ∞) excluding X = 0.
To sketch a graph of F, we use the information we have gathered:
1. The function F is increasing in the interval (-∞, 8).
2. Relative max point at (8, 16).
3. There are no relative min points.
4. F is concave down for the interval (-∞, ∞) excluding X = 0.
Based on this information, we can sketch the graph accordingly.
I read your statement as
f(x) = 12x^(2/3) - 4x
then
f '(x) = 8x^(-1/3) - 4
= 0 for max/min
8/x^(1/3) = 4
4x^(1/3) = 8
x^(1/3) = 2
x = 8
then f(8) = 12(8)^(2/3) - 4(8)
= 48 - 32
= 16
Can you take it from here?
remember that the function is concave up when the second derivative is positive and
concave down .....