Consider the function y = 3x5 – 25x3 + 60x + 1. Use the first or second derivative test to test the critical points. How many relative maxima did you find?
I got 2 as the relative maxima, is this done correctly?
Second derivative test to test the critical points
I did most of this in the previous post
Here is a picture of your graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+3x%5E5+%E2%80%93+25x%5E3+%2B+60x+%2B+1
To test the critical points using the first or second derivative test, you need to find the critical points first.
The critical points occur where the derivative of the function is equal to zero or undefined. Let's find the derivative of the given function y = 3x^5 – 25x^3 + 60x + 1:
y' = 15x^4 - 75x^2 + 60
To find the critical points, set the derivative equal to zero and solve for x:
15x^4 - 75x^2 + 60 = 0
Now, we can solve this equation. Factoring out the common factor of 15:
15(x^4 - 5x^2 + 4) = 0
Using factoring, we have:
15(x^2 - 4)(x^2 - 1) = 0
Now, set each factor equal to zero and solve for x:
x^2 - 4 = 0 or x^2 - 1 = 0
Solving these equations, we get:
x = ±2 or x = ±1
So, we have four critical points: x = -2, x = -1, x = 1, and x = 2.
Now, let's test each of these critical points using the first or second derivative test. This test will help us determine whether each critical point is a relative maximum, relative minimum, or neither.
We already have the first derivative, y' = 15x^4 - 75x^2 + 60. Let's find the second derivative:
y'' = 60x^3 - 150x
Now, let's evaluate the second derivative at each critical point to determine their nature:
1. For x = -2:
y''(-2) = 60(-2)^3 - 150(-2) = -120 - (-300) = 180
Since the second derivative is positive, we have a relative minimum at x = -2.
2. For x = -1:
y''(-1) = 60(-1)^3 - 150(-1) = -60 + 150 = 90
Since the second derivative is positive, we have a relative minimum at x = -1.
3. For x = 1:
y''(1) = 60(1)^3 - 150(1) = 60 - 150 = -90
Since the second derivative is negative, we have a relative maximum at x = 1.
4. For x = 2:
y''(2) = 60(2)^3 - 150(2) = 480 - 300 = 180
Since the second derivative is positive, we have a relative minimum at x = 2.
From the above analysis, we found two relative maxima at x = -2 and x = 2. Therefore, your conclusion that there are 2 relative maxima is correct.
To determine the relative extrema of a function using the first or second derivative test, we need to find the critical points of the function and then analyze the concavity of the function at those points.
To start, let's first find the critical points by finding where the derivative of the function is equal to zero or undefined. The derivative of the given function y = 3x^5 - 25x^3 + 60x + 1 can be obtained by applying the power rule:
y' = 15x^4 - 75x^2 + 60
Next, we set the derivative equal to zero and solve for x:
15x^4 - 75x^2 + 60 = 0
Factoring out the common factor of 15, we get:
15(x^4 - 5x^2 + 4) = 0
Now, we can solve the quadratic equation inside the parentheses:
x^4 - 5x^2 + 4 = 0
Factorizing the quadratic, we have:
(x^2 - 1)(x^2 - 4) = 0
(x - 1)(x + 1)(x - 2)(x + 2) = 0
This gives us four critical points: x = -2, -1, 1, and 2.
Now, we can use the second derivative test to determine the types of relative extrema.
The second derivative of the given function y = 3x^5 - 25x^3 + 60x + 1 can be obtained by differentiating the derivative we found earlier:
y'' = 60x^3 - 150x
We plug in the critical points obtained earlier into the second derivative to determine the concavity:
For x = -2:
y''(-2) = 60(-2)^3 - 150(-2) = -480 + 300 = -180
For x = -1:
y''(-1) = 60(-1)^3 - 150(-1) = -60 + 150 = 90
For x = 1:
y''(1) = 60(1)^3 - 150(1) = 60 - 150 = -90
For x = 2:
y''(2) = 60(2)^3 - 150(2) = 480 - 300 = 180
Since the second derivative changes sign, we can conclude that there is a relative maximum at x = -1 and a relative minimum at x = 1.
Therefore, there is one relative maximum and one relative minimum, totaling two relative extrema. So, your answer of 2 as the number of relative maxima is incorrect.