Find the intervals of increase and decrease algebraically for:
Y = −4x³+15x²+ 18x + 3
The function is increasing when the first derivative is positive, and
it is decreasing .... negative
dy/dx = -12x^2 + 30x + 18
consider the graph of y = -12x^2 + 30x +18
it is a parabola opening down with x-intercepts at x = 3 and x = -1/2
So the function is
increasing for -1/2 < x < 3
decreasing for x<-1/2 or x > 3
confirmation:https://www.wolframalpha.com/input/?i=+y+%3D+%E2%88%924x%5E3%2B15x%5E2+%2B+18x+%2B+3+from+-2+to+6
Thank you
To find the intervals of increase and decrease for the given function algebraically, you would need to follow these steps:
1. Calculate the first derivative of the function Y with respect to x. Let's call this function Y'(x).
2. Set Y'(x) equal to zero and solve for x to find the critical points.
3. Make a number line and plot the critical points on it.
4. Test a value in each interval between the critical points and determine whether Y'(x) is positive or negative in that interval. If Y'(x) is positive, the function is increasing; if negative, the function is decreasing.
Let's go through the steps for the given function Y = -4x³ + 15x² + 18x + 3:
Step 1:
Find the derivative of Y(x):
Y'(x) = -12x² + 30x + 18
Step 2:
Set Y'(x) = 0 and solve for x:
-12x² + 30x + 18 = 0
You can either factor this quadratic equation or use the quadratic formula to solve for x. Factoring does not seem straightforward, so let's use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
For our equation, a = -12, b = 30, and c = 18. Plugging these values into the quadratic formula:
x = (-30 ± √(30² - 4(-12)(18))) / (2(-12))
x = (-30 ± √(900 + 864)) / (-24)
x = (-30 ± √1764) / (-24)
x = (-30 ± 42) / (-24)
This gives us two possible values for x: x = 1 and x = -3/2.
Step 3:
Make a number line and plot the critical points:
-∞ ----- (-3/2) ----- 1 ----- ∞
Step 4:
Test a value within each interval to determine the sign of Y'(x).
- Test x = -4: Y'(-4) = -12(-4)² + 30(-4) + 18 = -192 + (-120) + 18 = -294
Since Y'(-4) is negative, the function is decreasing in the interval (-∞, -3/2).
- Test x = 0: Y'(0) = -12(0)² + 30(0) + 18 = 18
Since Y'(0) is positive, the function is increasing in the interval (-3/2, 1).
- Test x = 2: Y'(2) = -12(2)² + 30(2) + 18 = -48 + 60 + 18 = 30
Since Y'(2) is positive, the function is increasing in the interval (1, ∞).
Therefore, the intervals of increase and decrease for Y = -4x³ + 15x² + 18x + 3 are:
Decreasing: (-∞, -3/2)
Increasing: (-3/2, 1), (1, ∞)