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?

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So I put the derivative of 15x^4-75x^2+60 into wolfram-alpha and it got four critial points on the graph, but they're all equally leveled...so does mean there is no relative maxima or that there are four of them?

To test the critical points using the first or second derivative test, we first need to find the critical points by setting the first derivative equal to zero and solving for x.

The first derivative of the function y = 3x^5 - 25x^3 + 60x + 1 is given by:
dy/dx = 15x^4 - 75x^2 + 60

Setting dy/dx = 0, we have:
15x^4 - 75x^2 + 60 = 0

Factorizing the equation, we get:
3(x^4 - 5x^2 + 4) = 0

Next, we solve for x:
x^4 - 5x^2 + 4 = 0

This equation can be factored as a quadratic equation:
(x^2 - 1)(x^2 - 4) = 0

Simplifying further, we have:
(x + 1)(x - 1)(x + 2)(x - 2) = 0

So the critical points are x = -2, -1, 1, and 2.

To determine the number of relative maxima, we use the second derivative test. We need to find the second derivative of the function.
The second derivative of y = 3x^5 - 25x^3 + 60x + 1 is given by:

d^2y/dx^2 = 60x^3 - 150x

Next, we substitute the critical points into the second derivative to check the sign of the second derivative at each critical point.

At x = -2:
d^2y/dx^2 = 60(-2)^3 - 150(-2) = -480 - (-300) = -480 + 300 = -180 (negative)

At x = -1:
d^2y/dx^2 = 60(-1)^3 - 150(-1) = -60 - (-150) = -60 + 150 = 90 (positive)

At x = 1:
d^2y/dx^2 = 60(1)^3 - 150(1) = 60 - 150 = -90 (negative)

At x = 2:
d^2y/dx^2 = 60(2)^3 - 150(2) = 480 - 300 = 180 (positive)

From the second derivative test, we observe that the signs of the second derivative alternate between positive and negative at the critical points.

This means that at x = -2 and x = 1, there are relative maxima. Therefore, we have 2 relative maxima.

So the correct answer is 2.

To test the critical points using the first or second derivative test, we need to find the critical points first. Critical points occur when the first derivative is equal to zero or undefined.

Let's find the first derivative of the given function y = 3x^5 – 25x^3 + 60x + 1:

dy/dx = 15x^4 - 75x^2 + 60

Now, let's find the critical points by setting the first derivative equal to zero and solving for x:

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

We can simplify this equation by factoring out 15:

15(x^4 - 5x^2 + 4) = 0

Now, we have a quadratic equation x^4 - 5x^2 + 4 = 0. This equation can be factored as (x^2 - 4)(x^2 - 1) = 0.

Setting each factor equal to zero, we get:

x^2 - 4 = 0 and x^2 - 1 = 0

Solving these two equations, we find four critical points:

x = -2, x = 2, x = -1, and x = 1

Now, let's use the first or second derivative test to determine the nature of these critical points.

Using the first derivative test, we evaluate the sign of the first derivative (+) or (-) in intervals around each critical point. If the sign changes from positive to negative, the critical point is a relative maximum. If the sign changes from negative to positive, the critical point is a relative minimum.

For the interval x < -2:
Choose x = -3
dy/dx = 15(-3)^4 - 75(-3)^2 + 60
dy/dx = 2700 - 675 + 60
dy/dx = 2085 > 0 (positive)

For the interval -2 < x < -1:
Choose x = -1.5
dy/dx = 15(-1.5)^4 - 75(-1.5)^2 + 60
dy/dx = 67.5 + 168.75 + 60
dy/dx = 296.25 > 0 (positive)

For the interval -1 < x < 1:
Choose x = 0
dy/dx = 15(0)^4 - 75(0)^2 + 60
dy/dx = 60 > 0 (positive)

For the interval 1 < x < 2:
Choose x = 1.5
dy/dx = 15(1.5)^4 - 75(1.5)^2 + 60
dy/dx = 67.5 - 168.75 + 60
dy/dx = -41.25 < 0 (negative)

For the interval x > 2:
Choose x = 3
dy/dx = 15(3)^4 - 75(3)^2 + 60
dy/dx = 2430 - 675 + 60
dy/dx = 1815 > 0 (positive)

Now, by examining the sign changes, we can determine the relative maxima:

-2 to -1: No sign change (not a relative maxima)
-1 to 1: No sign change (not a relative maxima)
1 to 2: Sign changes from positive to negative (relative maxima)

Based on the analysis, we have found one relative maximum. Therefore, the correct answer is 1.