The directions say to use a grapking utility to approximate (to two-decimal-place accuracy) any relative minimum or maximum values of the function:
y=x^3 - 6x^2 + 15
Can someone please help me because my graphing calculator is not working!!!
y' = 3 x^2 - 12 x
y' = 0 at max or min
3x (x-4) = 0
x = 0 or x = 4 for slope horizontal
y" = 6 x - 12
at x = 0, y"=-12 so maximum
at x = 4, y" = +12 so minimum
then at x = 0, y = 15
and at x = 4, y = -17
So if you do not know calculus sketch the graph starting at lower left, rising to 15 at x = 0, dropping to -17 at x = 4, then rising up to upper right.
sketch some points right around x = 0 and right around x = 4
Of course, I'd be happy to help you approximate the relative minimum or maximum values of the function without using a graphing utility.
To find the relative minimum or maximum values of the given function y = x^3 - 6x^2 + 15, we can follow these steps:
1. Determine the critical points: Critical points occur where the derivative of the function is equal to zero or does not exist. In this case, we need to find where the derivative of y with respect to x is equal to zero.
Let's take the derivative of the given function:
dy/dx = 3x^2 - 12x
To find the critical points, we set the derivative equal to zero and solve for x:
3x^2 - 12x = 0
3x(x - 4) = 0
This equation gives us two potential critical points: x = 0 and x = 4.
2. Determine the nature of each critical point: To determine whether each critical point is a relative minimum or maximum, we can use the second derivative test. We need to evaluate the second derivative of the function y with respect to x.
Taking the second derivative of y, we get:
d^2y/dx^2 = 6x - 12
Now, let's evaluate the second derivative at each critical point:
For x = 0: d^2y/dx^2 = 6(0) - 12 = -12
For x = 4: d^2y/dx^2 = 6(4) - 12 = 12
According to the second derivative test:
- If the second derivative is positive, the point is a relative minimum.
- If the second derivative is negative, the point is a relative maximum.
In our case:
- For x = 0: Since the second derivative (-12) is negative, we have a relative maximum at this point.
- For x = 4: Since the second derivative (12) is positive, we have a relative minimum at this point.
3. Find the corresponding y-values: Plug the values of x into the original function y = x^3 - 6x^2 + 15 to find the corresponding y-values.
For x = 0, y = (0)^3 - 6(0)^2 + 15 = 15
For x = 4, y = (4)^3 - 6(4)^2 + 15 = -1
Therefore, the relative maximum occurs at the point (0, 15) and relative minimum occurs at the point (4, -1).
While graphing utilities provide a visual representation of functions, these steps allow you to find the relative minimum or maximum values of the function without relying on a calculator.