Find the range of values of x for which each of the following functions is decreasing
(x^2)/2-5x+1
x^3-(5/2)x^2-2x+1
To find the range of values of x for which a function is decreasing, we need to find the interval(s) where the derivative of the function is negative.
For the first function, f(x) = (x^2)/2 - 5x + 1, let's find its derivative:
f'(x) = (2x)/2 - 5 = x - 5
To find when f'(x) < 0:
x - 5 < 0
x < 5
So, the first function is decreasing for x < 5.
For the second function, f(x) = x^3 - (5/2)x^2 - 2x + 1, let's find its derivative:
f'(x) = 3x^2 - 5x - 2
To find when f'(x) < 0, we can use the quadratic formula:
x = (-(-5) ± sqrt((-5)^2 - 4(3)(-2))) / (2(3))
x = (5 ± sqrt(25 + 24)) / 6
x = (5 ± sqrt(49)) / 6
x = (5 ± 7) / 6
So, x = 2/3 or x = 4.
Therefore, the second function is decreasing for x < 2/3 and x > 4.
To find the range of values of x for which each function is decreasing, we need to find the critical points of each function and check whether the function is decreasing on the intervals between those critical points.
1. Function: f(x) = (x^2)/2 - 5x + 1
To find the critical points, we need to find the first derivative of f(x), set it equal to zero, and solve for x:
f'(x) = (d/dx)((x^2)/2) - (d/dx)(5x) + (d/dx)(1)
= x - 5
Setting f'(x) = 0, we have:
x - 5 = 0
Solving for x, we find that x = 5. Therefore, the critical point is x = 5.
Now, we need to determine if the function is decreasing to the left and right of x = 5. To do this, we can pick any value less than 5, say x = 0, and any value greater than 5, say x = 10, and evaluate f'(x) for both:
When x = 0, f'(0) = 0 - 5 = -5
When x = 10, f'(10) = 10 - 5 = 5
Since f'(x) is negative when x < 5 and positive when x > 5, we can conclude that the function f(x) is decreasing for x < 5.
2. Function: f(x) = x^3 - (5/2)x^2 - 2x + 1
Similarly, we need to find the first derivative of f(x), set it equal to zero, and solve for x:
f'(x) = 3x^2 - 5x - 2
Setting f'(x) = 0, we have:
3x^2 - 5x - 2 = 0
Solving for x using the quadratic formula, we find that x ≈ -0.5 and x ≈ 1.66. Therefore, the critical points are x ≈ -0.5 and x ≈ 1.66.
Again, we need to check if the function is decreasing to the left and right of these critical points. We can evaluate f'(x) for values less than -0.5, between -0.5 and 1.66, and greater than 1.66:
When x = -1, f'(-1) = 3(-1)^2 - 5(-1) - 2 = 2
When x = 0, f'(0) = 3(0)^2 - 5(0) - 2 = -2
When x = 2, f'(2) = 3(2)^2 - 5(2) - 2 = 6
Since f'(x) is positive when x < -0.5, negative when -0.5 < x < 1.66, and positive when x > 1.66, we can conclude that the function f(x) is decreasing for -0.5 < x < 1.66.
Therefore, the range of values of x for which each function is decreasing is:
1. Function (x^2)/2 - 5x + 1: x < 5
2. Function x^3 - (5/2)x^2 - 2x + 1: -0.5 < x < 1.66