1. Consider the function f(x) = X^(4/3) +4x^(1/3) on the interval -8 ≤ x ≤ 8.
a. Find the coordinate of all points at which the tangent to the curve is a horizontal line.
b. Find the coordinate s of all points at which the tangent to the curve is a vertical line.
c. Find the coordinates of the points at which the absolute maximum and absolute minimum occur.
a. To find the coordinates of all points at which the tangent to the curve is a horizontal line, we need to find the critical points of the function.
To find the critical points, we need to find where the derivative of the function equals zero. Let's find the derivative of f(x):
f'(x) = (4/3)x^(1/3) + (4/3)x^(-2/3)
Setting f'(x) equal to zero:
0 = (4/3)x^(1/3) + (4/3)x^(-2/3)
To simplify, let's multiply both sides of the equation by 3 to get rid of the fraction:
0 = 4x^(1/3) + 4x^(-2/3)
Now, let's solve for x. Subtracting 4x^(1/3) from both sides:
-4x^(1/3) = 4x^(-2/3)
Dividing both sides by -4x^(2/3):
x^(-2/3) = -x^(1/3)
Raising both sides to the power of -3:
x^2 = -1
Since x^2 = -1 has no real solutions, there are no points on the interval -8 ≤ x ≤ 8 where the tangent to the curve is a horizontal line.
b. To find the coordinates of all points at which the tangent to the curve is a vertical line, we need to find where the derivative of f(x) is undefined.
The derivative of f(x), f'(x), is already calculated above. To find where it is undefined, we need to find where the denominator of f'(x) is zero. In this case, the denominator is x^(2/3).
Setting x^(2/3) equal to zero:
0 = x^(2/3)
There are no real solutions to this equation, so there are no points on the interval -8 ≤ x ≤ 8 where the tangent to the curve is a vertical line.
c. To find the coordinates of the points at which the absolute maximum and absolute minimum occur, we need to find the critical points and endpoints of the interval.
As we found in part a, there are no critical points on the interval -8 ≤ x ≤ 8.
Now, let's evaluate the function at the endpoints of the interval:
f(-8) = (-8)^(4/3) + 4(-8)^(1/3)
= 8 + 4(-2)
= 8 - 8
= 0
f(8) = (8)^(4/3) + 4(8)^(1/3)
= 8 + 4(2)
= 8 + 8
= 16
Therefore, the absolute minimum occurs at (x, f(x)) = (-8, 0) and the absolute maximum occurs at (x, f(x)) = (8, 16).
To find the points where the tangent to the curve is a horizontal line, we need to find the derivative of the function and set it equal to zero. Let's start with part (a).
a. Find the points where the tangent to the curve is a horizontal line:
Step 1: Find the derivative of the function f(x).
To find the derivative of f(x) = x^(4/3) + 4x^(1/3), we can use the power rule.
f'(x) = (4/3)x^(1/3) + (4/3)x^(-2/3)
Step 2: Set the derivative equal to zero and solve for x:
(4/3)x^(1/3) + (4/3)x^(-2/3) = 0
Multiply both sides of the equation by 3 to get rid of the denominators:
4x^(1/3) + 4x^(-2/3) = 0
Step 3: Solve for x.
We can simplify the equation further by factoring out 4:
4(x^(1/3) + x^(-2/3)) = 0
Now we have two cases to consider:
Case 1: x^(1/3) + x^(-2/3) = 0
This equation does not have any solutions since the sum of two positive numbers cannot be zero.
Case 2: x = 0
Therefore, the only solution is x = 0.
So, the coordinate of the point where the tangent to the curve is a horizontal line is (0, f(0)).
Now let's move on to part (b).
b. Find the points where the tangent to the curve is a vertical line:
To find the points where the tangent to the curve is a vertical line, we need to look for vertical tangent lines, which occur when the derivative is undefined.
Step 1: Find the derivative of f(x).
We already found f'(x) in part (a), which is:
f'(x) = (4/3)x^(1/3) + (4/3)x^(-2/3)
Step 2: Set the derivative equal to undefined.
Vertical tangent lines occur when the derivative is undefined, which happens when the denominator of the derivative is equal to zero. In this case, when x^(-2/3) is equal to zero.
x^(-2/3) = 0
To solve this equation, we can raise both sides to the -3/2 power:
(x^(-2/3))^(-3/2) = 0^(-3/2)
x = 0
Therefore, the only point where the tangent to the curve is a vertical line is x = 0.
Moving on to part (c).
c. Find the coordinates of the points where the absolute maximum and absolute minimum occur:
To find the points where the absolute maximum and absolute minimum occur, we need to find the critical points of the function f(x) and evaluate the function at these points.
Step 1: Find the critical points of f(x).
The critical points occur when the derivative is equal to zero or undefined. From part (a) and part (b), we have already found the critical point to be x = 0.
Step 2: Evaluate the function at the critical points.
f(x) = x^(4/3) + 4x^(1/3)
When x = 0:
f(0) = 0^(4/3) + 4(0)^(1/3)
f(0) = 0 + 0
f(0) = 0
Therefore, the coordinate of the point where the absolute maximum and absolute minimum occur is (0, 0).
To summarize:
a. The coordinate of the point where the tangent to the curve is a horizontal line is (0, f(0)).
b. The coordinate of the point where the tangent to the curve is a vertical line is x = 0.
c. The coordinate of the point where the absolute maximum and absolute minimum occur is (0, 0).