Critical Points of f(x): -2x^3 + 39x^2 - 180x + 1

In summary, the function f(x) is -2x^3 + 39x^2 - 180x + 1 and the task is to list all critical points and determine where it is increasing and decreasing. The correct critical points are at -2 and 15, with a negative slope before -2, positive between -2 and 15, and negative after 15. However, there was a mistake in solving the quadratic equation, and the correct roots are 3 and 10. The second derivative rule was used to determine that -2 is a local minimum and 15 is a local maximum. It is unclear if there are any additional critical points.
  • #1
smith5029
8
0
hi guys I'm stuck on this, i keep thinking i have the right answeres but i can't get them.
the function is f(x): -2x^3 + 39x^2 - 180x + 1
i need to list all of the critical points, and hten indicate where it is increasing and where it is decreasing.

I set the derivative -6x^2 + 78x - 180 = 0 and then solved for x. this is correct right. by doing so i ended up with critical points at -2 and 15. Then made a number line and figured that the slope before -2 was neg, b/w -2 and 15 it was pos, and after 15 it was neg. so my interval of increasing was (-2,15) and the interval of decreasing was (-inf, -2) U (15,inf) where am i going wrong - PLEASE HELP ME
 
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  • #2
Did u compute the second derivative,too...?To find the nature of those 2 points,u must evaluate the second derivative in those points.

Daniel.
 
  • #3
You made a mistake in solving the quadratic. The roots are 3 and 15.
 
  • #4
Yes i determined the second derivative as -12x + 78. Then using the second derivative rule plugged in 15, and -2 telling me that -2 was a local min and 15 was a local max. DO you know if 15, -2 are the only critical points in my problem?
 
  • #5
i don't see how u got 3 and 15 ??
 
  • #7
-6x^2 + 78x - 180 = 0

x^2 -13x + 30 = 0
x=(13+/-sqrt(169-4*1*30))/2
x=(13+/-7)/2

x=20/2 = 10
x=6/2= 3
 

1. What are critical points?

Critical points are points on a function where the derivative is equal to zero or does not exist. These points are important because they can indicate where the function reaches a maximum, minimum, or inflection point.

2. How do you find the critical points of a function?

To find the critical points of a function, set the derivative equal to zero and solve for the variables. Then, plug these values back into the original function to determine the x-coordinates of the critical points.

3. What are the types of critical points?

The types of critical points are local minimum points, local maximum points, and inflection points. Local minimum points are where the function reaches its lowest value in a specific interval, while local maximum points are where the function reaches its highest value in a specific interval. Inflection points are where the concavity of the function changes.

4. How do critical points affect the behavior of a function?

Critical points can determine the behavior of a function by indicating where the function changes from increasing to decreasing, or vice versa. They also help identify where the function reaches its extreme values or changes concavity.

5. What is the significance of critical points in real-world applications?

Critical points are significant in real-world applications because they can help optimize a function's behavior. For example, in economics, critical points can determine the maximum profit or minimum cost for a company. In physics, critical points can indicate where an object reaches its maximum speed or acceleration.

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