Can someone please show steps as I am having a hard time understanding. The position (in meters) of a particle is modeled by y=x^3-12.5x^2+51x-67.5, where xis in seconds. Find the max, min and point(s) of inflection.
Reiny gave you some tips. Where do you get stuck?
we got a few examples in class but not understanding it at all
Oh, please. There must be something you understand by this time in the course.
You know that max/min occurs where y'=0. Do you know how to find y'?
yes, i pretty sure i got it as my answer mathes the back of the book
Well, here goes. If you really are totally lost, you'll probably have much better progress talking to someone in person.
y=x^3-12.5x^2+51x-67.5
Now, you know that y' is the slope of the tangent line to the curve at (x,y). When that tangent line is horizontal, the curve is either at a max or a min. Think of a hill or valley -- at the top of the hill, you stop rising (slope >0), and start falling (slope <0)
y' = 3x^2 -25x + 51
to find y'=0, solve for x
0 = 3x^2 - 25x + 51
x = 1/6 (25 ±√13)
x = 3.57 or 4.77
Now, from your (presumed) general knowledge of the shape of cubics, you should be able to surmise that the first value is where y reaches a max, the second where it reaches a min.
A point of inflection occurs when the slope reaches a max or min. That's where the curve changes from concave up to concave down. It is where y''=0 and y' ≠ 0.
y'' = 6x - 25
y'' = 0 when x = 25/6 = 4.17
plug in those x-values to find the actual y-coordinates of the points.
To find the maximum, minimum, and point(s) of inflection of a graph, we need to analyze the critical points and the concavity of the function.
Step 1: Find the derivative of the position function (y) with respect to time (x) to obtain the velocity function (v).
Given: y = x^3 - 12.5x^2 + 51x - 67.5
To find the derivative, we use the power rule:
dy/dx = 3x^2 - 25x + 51
Step 2: Find the critical points of the velocity function.
The critical points occur when the velocity is equal to zero, so we solve the equation 3x^2 - 25x + 51 = 0 for x.
Using the quadratic formula, x = (-b ± √(b^2 - 4ac))/(2a), we get:
x = (25 ± √(25^2 - 4*3*51))/(2*3)
x = (25 ± √(625 - 612))/6
x = (25 ± √13)/6
These are the two critical points.
Step 3: Determine the concavity of the function.
To find the points of inflection, we need to determine where the concavity changes. This occurs when the second derivative of the position function changes sign.
The second derivative is found by taking the derivative of the velocity function:
d^2y/dx^2 = 6x - 25
Step 4: Find the points of inflection.
We set the second derivative equal to zero and solve for x:
6x - 25 = 0
x = 25/6
This is the point of inflection.
Step 5: Determine the nature of the critical points.
To determine if the critical points are a maximum or minimum, we evaluate the second derivative at each of the critical points.
For x = (25 + √13)/6, d^2y/dx^2 = 6((25 + √13)/6) - 25 = √13 - 1, which is positive. This means this critical point corresponds to a minimum.
For x = (25 - √13)/6, d^2y/dx^2 = 6((25 - √13)/6) - 25 = -√13 - 1, which is negative. This means this critical point corresponds to a maximum.
Step 6: Find the coordinates of the maximum, minimum, and point(s) of inflection.
To find the y-coordinates of each point, substitute the x-values into the original function:
For x = (25 + √13)/6, y = ((25 + √13)/6)^3 - 12.5((25 + √13)/6)^2 + 51((25 + √13)/6) - 67.5
For x = (25 - √13)/6, y = ((25 - √13)/6)^3 - 12.5((25 - √13)/6)^2 + 51((25 - √13)/6) - 67.5
For x = 25/6, y = (25/6)^3 - 12.5(25/6)^2 + 51(25/6) - 67.5
These calculations will give you the coordinates of the maximum, minimum, and point(s) of inflection.