Consider the function y = 3x5 – 25x3 + 60x + 1. Find the critical points of this function.

So I got 1 and 2 as the critical points..but I feel like I did something wrong, is this right?

So the points would be -1,-2,1,2?

it is -2, -1, 1, 2

I am sure you meant:

y = 3x^5 - 25x^3 + 60x + 1
dy/dx = 15x^4 - 75x^2 + 60
= 0 for your critical values
3x^4 - 15x^2 + 12 = 0
x^4 - 5x^2 + 4 = 0
(x^2 - 1)(x^2 - 4) = 0
x^2 = 1 or x^2 = 4
x = ± 1 OR x = ± 2

sub in those values of x in the original to get the mataching y values of the critical points.

Some would consider the inflection points in the list of critical points

y'' = 60x^3 - 150x
= 0 for points of inflection
6x^3 - 15x = 0
2x^3 - 5x = 0
x(2x^2 - 5) = 0
x = 0 or x = ± 5/2
continue to find the points of inflection

Well, when it comes to critical points, it's all about finding where the function has turning points or changes in concavity.

To find critical points, we'll need to find the derivative of the given function. So let's differentiate y = 3x^5 – 25x^3 + 60x + 1.

Taking the derivative, we get: y' = 15x^4 - 75x^2 + 60.

Now, let's set this derivative equal to zero and solve for x:

15x^4 - 75x^2 + 60 = 0.

Hmm, this seems like a tough equation to solve algebraically. We might need to use numerical methods or a graphing calculator to find the exact points.

But just for the fun of it, let's try factoring. Can't hurt to try, right?

Factoring out a common factor of 15, we have:
15(x^4 - 5x^2 + 4) = 0.

And voila! We can factor the quadratic inside the parentheses as:
15(x^2 - 4)(x^2 - 1) = 0.

So, we have three possible critical points:
1. When x^2 - 4 = 0, we get x = ±2.
2. When x^2 - 1 = 0, we get x = ±1.

Therefore, it appears that the critical points of the function are x = -2, -1, 1, and 2.

Now that we've got the critical points, go rock those concavities!

To find the critical points of a function, you need to determine where its derivative is equal to zero or undefined. The critical points correspond to potential local extrema (maximum or minimum points) or points of inflection (where the concavity of the function changes).

To find the critical points of the function y = 3x^5 – 25x^3 + 60x + 1, we need to first take its derivative with respect to x. So, let's differentiate the function.

dy/dx = d/dx(3x^5) – d/dx(25x^3) + d/dx(60x) + d/dx(1)

To find the derivative of each term, we use the power rule:

dy/dx = 15x^4 – 75x^2 + 60

Now, we equate the derivative to zero and solve for x:

15x^4 – 75x^2 + 60 = 0

This can be factored as:

3(5x^4 – 25x^2 + 20) = 0

Factoring further:

3(x^2 – 1)(5x^2 – 20) = 0

Setting each factor to zero:

x^2 – 1 = 0 --> (1)
5x^2 – 20 = 0 --> (2)

Now, let's solve equation (1):

x^2 – 1 = 0

This gives two possible solutions:

x^2 = 1

Taking the square root of both sides:

x = ±1

So, ±1 are potential critical points.

Now, let's solve equation (2):

5x^2 – 20 = 0

Dividing both sides by 5:

x^2 – 4 = 0

This gives:

x^2 = 4

Taking the square root of both sides:

x = ±2

So, ±2 are also potential critical points.

To summarize, the potential critical points of the function y = 3x^5 – 25x^3 + 60x + 1 are x = ±1 and x = ±2.

Therefore, you were correct in identifying 1 and 2 as potential critical points.

no, points consist of ordered pairs of two numbers.

±1 and ±2 are the x's of the points.

I told you what to do with those x's above, please read it.