Consider the function y = 3x^5 – 25x^3 + 60x + 1 Use the first or second derivative test to test the critical points. How many relative maxima did you find?
I got this as the derivative; 15 (4-5 x^2+x^4)
And I got 4 relative maxima.
Come on. There's no way to have two relative maxima without either a minimum or an inflection point in between them.
y' = 15(x^4-5x^2+1)
y" = 30x(2x^2-5)
y" =0 when x = ±√(5/2)
So, f is concave
down in (-∞,-√(5/2))
up in (-√(5/2),0)
down in (0,√(5/2))
up in (√(5/2),+∞)
So, you have two minima and 2 maxima
Check it out at
http://www.wolframalpha.com/input/?i=3x^5+%E2%80%93+25x^3+%2B+60x+%2B+1
Well, it seems like you've reached critical point junction! Let's put on our detective hats and find out if you've got the right solution.
To test the critical points, we need to examine the sign changes in the first derivative. In this case, you mentioned the first derivative is 15(4-5x^2+x^4).
To determine relative maxima, we need to find the points where the first derivative changes from positive to negative. In simpler terms, we're looking for those points where the function goes from going up (positive slope) to going down (negative slope).
Now, let's put those brain wrinkles to work and solve this mystery. To find the relative maxima, we'll need to find the points where the first derivative changes from positive to negative.
You did a great job finding the first derivative, but let me clarify something. The derivative should be 15(4-5x^2+x^4). So close, but it appears you added an extra x when finding the derivative.
Now, all we have to do is set the first derivative equal to zero and find the values of x where it changes the sign. By doing this, we can locate the relative maxima. The number of relative maxima will be equal to the number of sign changes.
So, I hate to burst your bubble, but there won't be four relative maxima. There will actually be an odd number of relative maxima, depending on the number of sign changes.
Keep at it, detective! Don't let this setback deflate your confidence, and keep on solving!
To test the critical points using the first or second derivative test, we need to find the critical points first. The critical points occur where the derivative is equal to zero or undefined.
The derivative of the function y = 3x^5 – 25x^3 + 60x + 1 is given as 15(4-5x^2+x^4).
To find the critical points, we need to solve the equation 15(4-5x^2+x^4) = 0.
Setting the derivative equal to zero gives us:
15(4-5x^2+x^4) = 0
Simplifying the equation:
4 - 5x^2 + x^4 = 0
This is a quadratic equation in terms of x^2. Solving this equation gives us the potential critical points.
Factoring out x^2, we have:
x^2(x^2-5)+4=0
Setting each factor equal to zero, we get:
x^2 = 0 or x^2-5 = 0
Solving for x, we find:
x = 0, -√5, √5
Now we have three critical points: x = 0, x = -√5, and x = √5.
To determine the nature of each critical point, we need to use the second derivative.
The second derivative of the function is given by:
f''(x) = 30x(2-x^2)
We need to evaluate the second derivative at each critical point to determine whether they correspond to relative maxima, minima, or neither.
Let's substitute each critical point into the second derivative:
At x = 0: f''(0) = 30(0)(2-0^2) = 0. Since the second derivative is equal to zero, we cannot determine the nature of this critical point using the second derivative test.
At x = -√5: f''(-√5) = 30(-√5)(2-(-√5)^2) = -30√5. Since the result is negative, the critical point x = -√5 corresponds to a relative maximum.
At x = √5: f''(√5) = 30(√5)(2-(√5)^2) = 30√5. Since the result is positive, the critical point x = √5 corresponds to a relative minimum.
Therefore, after testing the critical points using the first and second derivative test, we have two relative extrema. One is a relative maximum at x = -√5 and the other is a relative minimum at x = √5.
To find the relative maxima of a function using the first or second derivative test, we need to follow these steps:
1. Find the first derivative of the function. In this case, you correctly found the derivative as:
y' = 15(4 - 5x^2 + x^4)
2. Set the derivative equal to zero to find the critical points:
15(4 - 5x^2 + x^4) = 0
Simplifying the equation, we get:
4 - 5x^2 + x^4 = 0
This is a quadratic equation in terms of x^2. Solving it will give us the critical points.
Factorizing the equation, we get:
(x^2 - 1)(x^2 - 4) = 0
Solving each factor individually:
x^2 - 1 = 0: This gives us two solutions, x = -1 and x = 1.
x^2 - 4 = 0: This gives us two additional solutions, x = -2 and x = 2.
Therefore, we have four critical points: x = -2, x = -1, x = 1, and x = 2.
3. Now, we need to determine whether these critical points are relative maxima. We can use the second derivative test to determine this. Let's find the second derivative first.
Taking the derivative of the first derivative, we get:
y'' = d(15(4 - 5x^2 + x^4))/dx = -60x + 30x^3
4. Evaluate the second derivative at each critical point we found:
For x = -2:
y''(-2) = -60(-2) + 30(-2)^3 = 120 - 240 = -120
For x = -1:
y''(-1) = -60(-1) + 30(-1)^3 = 60 - 30 = 30
For x = 1:
y''(1) = -60(1) + 30(1)^3 = -60 + 30 = -30
For x = 2:
y''(2) = -60(2) + 30(2)^3 = -120 + 240 = 120
5. Now, analyze the signs of the second derivatives to determine whether each critical point corresponds to a relative maximum.
- For x = -2: y''(-2) is negative, indicating a concave-down shape. So, x = -2 corresponds to a relative maximum.
- For x = -1: y''(-1) is positive, indicating a concave-up shape. So, x = -1 corresponds to neither a relative minimum nor a relative maximum.
- For x = 1: y''(1) is negative, indicating a concave-down shape. So, x = 1 corresponds to a relative maximum.
- For x = 2: y''(2) is positive, indicating a concave-up shape. So, x = 2 corresponds to neither a relative minimum nor a relative maximum.
6. Based on the analysis, we have two relative maxima: one at x = -2 and another at x = 1.
Therefore, the correct answer is: there are two relative maxima.