How do you find the intervals of increasing and decreasing given #y=-x^3+2x^2+2#?

Question

Christopher Agresta · Accepted Answer

Well, the slope of the equation's graph at each point is represented by the first derivative.

#dy/dx = -3x^2 + 4x#

Wherever this is greater than zero, the function is increasing.

X can be factored out:

#dy/dx = x(-3x + 4)#

...and this tells us that the slope of the graph of this eq. is zero when x = 0 and #-3x + 4 = 0#, or #x = 4/3#

These prove to be the interval's endpoints where the function is increasing, but you can further analyze the data:

#dy/dx = x(-3x + 4) > 0#

Because a negative number times a negative number is positive, -3x is > 0, so the term (-3x + 4) must be > 0 if x is less than 0. Additionally, any positive number plus 4 is also positive.

Given that a negative number (keep in mind that x < 0) times a positive number is always negative, x (-3x+ 4) must therefore be NEGATIVE.

Thus, in the interval where x < 0, x is DECREASING.

Examine the term (-3x + 4) once more now.

if x is > 0, but less than 4/3, then this term is positive. If x is > 4/3, then #-3x + 4# is negative, so therefore the slope (#x(-3x + 4)#) will also be negative.

So, therefore 0 < x < 4/3 is the only interval where the original function #-x^3 + 2x^2 + 2# is increasing.

Alternatively, you could graph the function and use visual aids to identify the increasing interval.

Isaiah Allen · Answer

undefined To find the intervals of increasing and decreasing for the function y = -x^3 + 2x^2 + 2, you need to analyze the sign of its derivative.

1. Find the derivative of the function y = -x^3 + 2x^2 + 2.
y' = -3x^2 + 4x

2. Set the derivative equal to zero and solve for x to find critical points.
-3x^2 + 4x = 0
x(-3x + 4) = 0
x = 0 or x = 4/3

3. Create a sign chart for the derivative using the critical points.
Test a value in each interval created by the critical points (-∞, 0), (0, 4/3), and (4/3, ∞) into the derivative.

For x < 0, choose x = -1: 
y' = -3(-1)^2 + 4(-1) = 3 - 4 = -1 (negative)
For 0 < x < 4/3, choose x = 1: 
y' = -3(1)^2 + 4(1) = -3 + 4 = 1 (positive)
For x > 4/3, choose x = 2: 
y' = -3(2)^2 + 4(2) = -12 + 8 = -4 (negative)

4. Determine the intervals of increasing and decreasing.
The function is increasing when the derivative is positive and decreasing when the derivative is negative.

Interval of increasing: (0, 4/3)
Interval of decreasing: (-∞, 0) and (4/3, ∞)