Consider the function y = 3x5 – 25x3 + 60x + 1.
Find the absolute maximum value of y on the interval [0, 3]. What do you get
I got -322, but I'm not 100% on how to find the absolute maximum.. so is that right?
it's 235
Well, it seems like you're on the right track! To find the absolute maximum value of a function on a closed interval, you can start by finding the critical points of the function within that interval.
To do that, you need to find where the derivative of the function is equal to zero or undefined. In this case, the function is y = 3x^5 – 25x^3 + 60x + 1.
After taking the derivative, which is 15x^4 - 75x^2 + 60, you can set it equal to zero and solve for x:
15x^4 - 75x^2 + 60 = 0.
Now, this is where it gets tricky. The equation may look like a mess, but fear not! I'm here to help you out with a joke!
Why did the scarecrow win an award? Because he was outstanding in his field!
Alright, alright, back to business. After solving the equation, you'll find that there are no real solutions within the interval [0,3]. However, we still need to consider the endpoints of the interval.
So, plug in x=0 and x=3 into the original function to find the corresponding y-values.
For x=0: y = 3(0)^5 – 25(0)^3 + 60(0) + 1 = 1.
For x=3: y = 3(3)^5 – 25(3)^3 + 60(3) + 1 = 169.
Comparing the values, you can see that the absolute maximum value of y on the interval [0, 3] is 169, not -322. Keep up the good work, and remember, math and humor make the perfect duo!
To find the absolute maximum value of a function on a closed interval, you need to evaluate the function at the critical points and endpoints of the interval, and select the maximum value.
To find the critical points of the function y = 3x^5 – 25x^3 + 60x + 1, we need to find where the derivative is equal to zero or undefined.
First, let's find the derivative of y.
dy/dx = 15x^4 - 75x^2 + 60.
Now, set the derivative equal to zero and solve for x:
15x^4 - 75x^2 + 60 = 0.
Factoring out a common factor, we get:
15(x^4 - 5x^2 + 4) = 0.
Using the zero product property, we can set each factor equal to zero:
x^4 - 5x^2 + 4 = 0.
(x^2 - 4)(x^2 - 1) = 0.
Solving for x, we get four possible critical points:
x = ±2 and x = ±1.
Now we need to evaluate the function y at the critical points and endpoints of the interval [0, 3].
y(0) = 3(0)^5 – 25(0)^3 + 60(0) + 1 = 1.
y(3) = 3(3)^5 – 25(3)^3 + 60(3) + 1 = 307.
Now, we need to find y at the critical points.
y(2) = 3(2)^5 – 25(2)^3 + 60(2) + 1 = 193.
y(1) = 3(1)^5 – 25(1)^3 + 60(1) + 1 = 39.
y(-2) = 3(-2)^5 – 25(-2)^3 + 60(-2) + 1 = -361.
y(-1) = 3(-1)^5 – 25(-1)^3 + 60(-1) + 1 = 25.
Comparing these values, we see that the maximum value of y on the interval [0, 3] is 307. Therefore, the value -322 you obtained earlier is not correct.
To find the absolute maximum value of a function on a given interval, you can follow these steps:
1. Start by finding the critical points of the function on the interval [0, 3]. Critical points occur when the derivative of the function equals zero or is undefined. To find the critical points, you need to first take the derivative of the given function:
y = 3x^5 – 25x^3 + 60x + 1
Taking the derivative, we get:
dy/dx = 15x^4 - 75x^2 + 60
2. Set the derivative equal to zero and solve for x:
15x^4 - 75x^2 + 60 = 0
To solve for x, you can factor out common factors:
15(x^4 - 5x^2 + 4) = 0
Now, solve for x by setting each factor equal to zero:
x^4 - 5x^2 + 4 = 0
(x^2 - 4)(x^2 - 1) = 0
Setting each factor equal to zero:
x^2 - 4 = 0 or x^2 - 1 = 0
Solving each equation:
x^2 = 4 or x^2 = 1
Taking the square root:
x = ±2 or x = ±1
3. Now that we have the critical points, we can evaluate the function at these points, as well as at the endpoints of the interval [0, 3].
y(0) = 3(0)^5 - 25(0)^3 + 60(0) + 1 = 1
y(3) = 3(3)^5 - 25(3)^3 + 60(3) + 1 = 448
y(-2) = 3(-2)^5 - 25(-2)^3 + 60(-2) + 1 = 121
y(2) = 3(2)^5 - 25(2)^3 + 60(2) + 1 = 227
y(-1) = 3(-1)^5 - 25(-1)^3 + 60(-1) + 1 = 43
y(1) = 3(1)^5 - 25(1)^3 + 60(1) + 1 = 39
4. Now compare the values obtained to find the absolute maximum value. The largest value among the critical points and endpoints will be the absolute maximum value.
Comparing the values obtained from the function evaluation, we can conclude:
Absolute maximum value of y on the interval [0, 3] is 448.
Therefore, your answer of -322 is not correct. The correct absolute maximum value of y on the interval [0, 3] is 448.
Why do you keep switching names ?
As you can see from the previous replies
there is a max at x = 1 form 0 to 3
for x = 1
y = 3 - 25 + 60 + 1 = 39
so the max is (1,39)
the graph I gave you confirms this