Find the range of values of x for which each of the following functions is increasing
(3/10)x^4-(4/5)x^3-3x^2+36/5x+1
1+3x-2x^2
x^3-12x
To find the range of values of x for which each function is increasing, we need to find the intervals on which the derivative of each function is positive.
1) For the function f(x) = (3/10)x^4 - (4/5)x^3 - 3x^2 + (36/5)x + 1, we can find the derivative f'(x) by taking the derivative of each term separately:
f'(x) = 4(3/10)x^3 - 3(4/5)x^2 - 2(3)x + (36/5)
Simplifying, we have:
f'(x) = (12/10)x^3 - (12/5)x^2 - 6x + (36/5)
We want to find the intervals where f'(x) > 0. To do this, we can find the critical points by setting f'(x) equal to zero and solving for x:
(12/10)x^3 - (12/5)x^2 - 6x + (36/5) = 0
Simplifying, we have:
2x^3 - 2x^2 - 5x + 6 = 0
By using synthetic division or polynomial long division, we can factor this equation to:
(x-1)(2x^2 - 4x - 6) = 0
Setting each factor equal to zero, we find the critical points:
x-1 = 0 --> x = 1
2x^2 - 4x - 6 = 0
Simplifying the quadratic equation, we have:
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
Setting each factor equal to zero, we find the critical points:
x-3 = 0 --> x = 3
x+1 = 0 --> x = -1
Now we can test the intervals created by these critical points:
For x < -1, we can pick a test point such as x = -2:
f'(-2) = (12/10)(-2)^3 - (12/5)(-2)^2 - 6(-2) + (36/5) = -16 + 12 + 12 + 7.2 = 15.2
Since f'(-2) > 0, this interval is part of the solution.
For -1 < x < 1, we can pick a test point such as x = 0:
f'(0) = (12/10)(0)^3 - (12/5)(0)^2 - 6(0) + (36/5) = 0 + 0 + 0 + 7.2 = 7.2
Since f'(0) > 0, this interval is also part of the solution.
For 1 < x < 3, we can pick a test point such as x = 2:
f'(2) = (12/10)(2)^3 - (12/5)(2)^2 - 6(2) + (36/5) = 19.2 - 9.6 - 12 + 7.2 = 4.8
Since f'(2) > 0, this interval is part of the solution.
For x > 3, we can pick a test point such as x = 4:
f'(4) = (12/10)(4)^3 - (12/5)(4)^2 - 6(4) + (36/5) = 76.8 - 38.4 - 24 + 7.2 = 21.6
Since f'(4) > 0, this interval is also part of the solution.
Therefore, the range of values of x for which f(x) = (3/10)x^4 - (4/5)x^3 - 3x^2 + (36/5)x + 1 is increasing is (-∞, -1) U (-1, 1) U (1, 3) U (3, ∞).
2) For the function f(x) = 1 + 3x - 2x^2, we can find the derivative f'(x) by taking the derivative of each term separately:
f'(x) = 3 - 4x
We want to find the intervals where f'(x) > 0. Setting f'(x) equal to zero, we have:
3 - 4x = 0
Solving for x, we get:
x = 3/4
Now we can test the intervals created by this critical point:
For x < 3/4, we can pick a test point such as x = 0:
f'(0) = 3 - 4(0) = 3
Since f'(0) > 0, this interval is part of the solution.
For x > 3/4, we can pick a test point such as x = 1:
f'(1) = 3 - 4(1) = -1
Since f'(1) < 0, this interval is not part of the solution.
Therefore, the range of values of x for which f(x) = 1 + 3x - 2x^2 is increasing is (-∞, 3/4).
3) For the function f(x) = x^3 - 12x, we can find the derivative f'(x) by taking the derivative of each term separately:
f'(x) = 3x^2 - 12
We want to find the intervals where f'(x) > 0. Setting f'(x) equal to zero, we have:
3x^2 - 12 = 0
Simplifying, we have:
x^2 - 4 = 0
(x-2)(x+2) = 0
Setting each factor equal to zero, we find the critical points:
x-2 = 0 --> x = 2
x+2 = 0 --> x = -2
Now we can test the intervals created by these critical points:
For x < -2, we can pick a test point such as x = -3:
f'(-3) = 3(-3)^2 - 12 = 27 - 12 = 15
Since f'(-3) > 0, this interval is part of the solution.
For -2 < x < 2, we can pick a test point such as x = 0:
f'(0) = 3(0)^2 - 12 = -12
Since f'(0) < 0, this interval is not part of the solution.
For x > 2, we can pick a test point such as x = 3:
f'(3) = 3(3)^2 - 12 = 27 - 12 = 15
Since f'(3) > 0, this interval is also part of the solution.
Therefore, the range of values of x for which f(x) = x^3 - 12x is increasing is (-∞, -2) U (2, ∞).
To find the range of values for which each of these functions is increasing, we need to find the intervals where the derivative of the function is positive.
1. Function: (3/10)x^4 - (4/5)x^3 - 3x^2 + (36/5)x + 1
Take the derivative of the function to find its critical points:
f'(x) = (12/10)x^3 - (12/5)x^2 - 6x + (36/5)
Setting f'(x) > 0, we can solve for x:
(12/10)x^3 - (12/5)x^2 - 6x + (36/5) > 0
Simplifying the inequality, we have:
(6/5)x^3 - (6/5)x^2 - 3x + (9/5) > 0
Now, we need to solve this inequality to find the range of x values for which the original function is increasing. However, the inequality doesn't factor easily, so we can use a graphing calculator or divide the inequality into smaller intervals to find the solution. Let's divide it into intervals based on the critical points of the function:
Interval 1: (-∞, x1)
Interval 2: (x1, x2)
Interval 3: (x2, x3)
Interval 4: (x3, +∞)
Find the x values for each interval by finding the roots of the derivative:
Let's solve the equation (12/10)x^3 - (12/5)x^2 - 6x + (36/5) = 0:
Interval 1: (-∞, x1)
No real roots exist.
Interval 2: (x1, x2)
x1 ≈ -3.599
Root: x2 ≈ -0.183
Interval 3: (x2, x3)
x2 ≈ -0.183
Root: x3 ≈ 5.782
Interval 4: (x3, +∞)
No real roots exist.
Therefore, the range of values for which the function (3/10)x^4 - (4/5)x^3 - 3x^2 + (36/5)x + 1 is increasing is approximately (-∞, -3.599) and (5.782, +∞).
2. Function: 1 + 3x - 2x^2
Take the derivative of the function to find its critical points:
f'(x) = 3 - 4x
Setting f'(x) > 0, we have:
3 - 4x > 0
Solving for x, we find:
x < 3/4
Therefore, the range of values for which the function 1 + 3x - 2x^2 is increasing is approximately (-∞, 3/4).
3. Function: x^3 - 12x
Take the derivative of the function to find its critical points:
f'(x) = 3x^2 - 12
Setting f'(x) > 0, we have:
3x^2 - 12 > 0
Simplifying, we find:
x^2 - 4 > 0
Factoring the quadratic, we have:
(x - 2)(x + 2) > 0
This inequality is true when x > 2 or x < -2.
Therefore, the range of values for which the function x^3 - 12x is increasing is approximately (-∞, -2) and (2, +∞).