I'm sorry, but the information provided in the question is not accurate or complete. To analyze the function h(x)=x-3x²+x correctly, we need to find the x-intercepts, local maximums, local minimums, and intervals of increase and decrease based on the correct calculations.
Let's break down the steps to find the x-intercepts, local maximums, local minimums, and intervals of increase and decrease for the function h(x)=x-3x²+x:
1. X-Intercepts:
To find the x-intercepts, we set h(x) equal to zero and solve for x:
0 = x - 3x² + x
0 = -3x² + 2x
0 = x(-3x + 2)
x = 0 (x-intercept at x = 0)
2. Local Maximums and Local Minimums:
To find the local maximums and minimums, we need to find the critical points of the function.
First, we find the derivative of h(x) to identify critical points:
h'(x) = 1 - 6x + 1
h'(x) = -6x + 2
Set h'(x) = 0 to find critical points:
0 = -6x + 2
6x = 2
x = 2/6
x = 1/3 (critical point)
Next, we determine the nature of the critical point by checking the second derivative:
h''(x) = -6
Since the second derivative is negative, the critical point x = 1/3 corresponds to a local maximum.
3. Function Behavior (Increase and Decrease):
To determine where the function is increasing or decreasing, we examine the sign of the first derivative:
When x < 1/3, h'(x) is positive, so the function is increasing.
When x > 1/3, h'(x) is negative, so the function is decreasing.
In conclusion:
- X-intercept: x = 0
- Local maximum at (1/3, h(1/3))
- Function is increasing for x < 1/3 and decreasing for x > 1/3
It's important to calculate these values accurately to understand the behavior of the function. If you have any further questions or need clarification, feel free to ask!
