How do you find the critical numbers and the end points of this function:
f(x)=2x^3-3x^2-12x+5
i calculated the critical numbers as x=-1 and x=2 is this right?!?!
great so far.
now find f(2) and f(-1) for the y values of your critical points.
unless there is a domain given for your x's , there are no "end points" for your function.
If by "ritical numbers" you mean where the function has a local maximum or minimum, that would be where
f'(x) = 6x^2 -6x -12 = 0
x^2 -x -2 = 0
(x+1)(x-2) = 0
yes, -1 and 2 are correct.
f(x) has no end points or absolute maximum or minimum. It goes to + and - infinity when x does the same
To find the critical numbers of a function, you need to find the values of x where the derivative of the function equals 0 or does not exist. The critical numbers of a function can be points of local maxima, local minima, or inflection points.
To find the critical numbers of the function f(x) = 2x^3 - 3x^2 - 12x + 5, you need to follow these steps:
1. Find the derivative of the function f(x) with respect to x. Let's call it f'(x).
f'(x) = 6x^2 - 6x - 12
2. Set f'(x) equal to 0 and solve for x to find the values where the derivative equals 0.
6x^2 - 6x - 12 = 0
To solve this quadratic equation, you can either factor it or use the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation:
x = (-(-6) ± √((-6)^2 - 4 * 6 * (-12))) / (2 * 6)
x = (6 ± √(36 + 288)) / 12
x = (6 ± √324) / 12
x = (6 ± 18) / 12
Simplifying further:
x1 = (6 + 18) / 12 = 24 / 12 = 2
x2 = (6 - 18) / 12 = -12 / 12 = -1
So, the critical numbers of the function f(x) = 2x^3 - 3x^2 - 12x + 5 are x = -1 and x = 2.
To check if these values are indeed critical numbers, you can substitute them back into the original function and observe the behavior around those points.
Additionally, to find the end points of the function, you need to determine the domain of the function. Since there are no restrictions on the domain of polynomial functions, the function f(x) = 2x^3 - 3x^2 - 12x + 5 is defined for all real numbers. Therefore, there are no specific end points to consider in this case.
So, your calculations are correct. The critical numbers of the function are x = -1 and x = 2.