Find the maximum and minimum values of the following function.
8x3 – 75x2 + 150x
To find the maximum and minimum values of this function, we need to first find the critical points.
Step 1: Take the derivative of the function with respect to x.
The derivative of the given function, f(x) = 8x^3 – 75x^2 + 150x, can be found using the power rule:
f'(x) = 24x^2 - 150x + 150
Step 2: Set the derivative equal to zero and solve for x to find the critical points.
24x^2 - 150x + 150 = 0
We can simplify this equation by dividing both sides by 6:
4x^2 - 25x + 25 = 0
Step 3: Use the quadratic formula to solve for x.
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 4, b = -25, and c = 25.
Using the quadratic formula, we get:
x = (-(-25) ± √((-25)^2 - 4(4)(25))) / (2 * 4)
Simplifying further, we have:
x = (25 ± √(625 - 400)) / 8
x = (25 ± √225) / 8
x = (25 ± 15) / 8
So, we have two possible critical points:
x1 = (25 + 15) / 8 = 40 / 8 = 5/2
x2 = (25 - 15) / 8 = 10 / 8 = 5/4
Step 4: Determine the type of critical point (maximum or minimum) by analyzing the second derivative.
To determine the type of critical point, we need to find the second derivative of the original function.
The second derivative, f''(x), can be found by taking the derivative of f'(x):
f''(x) = 48x - 150
Step 5: Substitute the critical points into the second derivative.
Substituting x1 = 5/2 and x2 = 5/4 into the second derivative, we get:
f''(5/2) = 48(5/2) - 150 = 120 - 150 = -30
f''(5/4) = 48(5/4) - 150 = 60 - 150 = -90
Step 6: Analyze the second derivative to determine the type of critical point.
If the second derivative is positive, the function has a minimum at that critical point. If the second derivative is negative, the function has a maximum at that critical point.
Since f''(5/2) = -30 < 0, the function has a maximum at x = 5/2.
Since f''(5/4) = -90 < 0, the function has a maximum at x = 5/4.
Step 7: Substitute the critical points and endpoints into the original function to find the maximum and minimum values.
To find the maximum and minimum values, we substitute the critical points and endpoints into the original function.
f(5/2) = 8(5/2)^3 – 75(5/2)^2 + 150(5/2)
= 8(125/8) – 75(25/4) + 150(5/2)
= 125 – 375 + 375
= 125
So, the maximum value of the function is 125.
Next, let's find the minimum value by evaluating the function at the endpoints.
f(0) = 8(0)^3 – 75(0)^2 + 150(0)
= 0 - 0 + 0
= 0
f(5) = 8(5)^3 – 75(5)^2 + 150(5)
= 8(125) – 75(25) + 750
= 1000 – 1875 + 750
= -125
Comparing the values, the minimum value of the function is -125.
Therefore, the maximum value is 125 and the minimum value is -125.
To find the maximum and minimum values of a function, we need to find the critical points. To do that, we need to find the derivative of the function and set it equal to zero.
Let's start by finding the derivative of the function:
f'(x) = 24x^2 - 150x + 150
Now we set the derivative equal to zero and solve for x to find the critical points:
24x^2 - 150x + 150 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not easy, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation, we get:
x = (-(-150) ± √((-150)^2 - 4(24)(150))) / (2(24))
x = (150 ± √(22500 - 14400)) / 48
x = (150 ± √8100) / 48
x = (150 ± 90) / 48
Now we have two possible solutions for x:
x1 = (150 + 90) / 48 = 240 / 48 = 5
x2 = (150 - 90) / 48 = 60 / 48 = 5/4
These are the critical points of the function. Now we need to evaluate the function at these points and the endpoints of the interval to find the maximum and minimum values.
Let's start by evaluating the function at the critical points:
f(5) = 8(5^3) – 75(5^2) + 150(5) = 1000 - 1875 + 750 = -125
f(5/4) = 8((5/4)^3) – 75((5/4)^2) + 150(5/4) = 200/8 - 375/16 + 375/4 = -240/16 = -15
Now let's evaluate the function at the endpoints of the interval. However, you have not given any interval, so we cannot determine the endpoints.
Therefore, the maximum value is -15 and the minimum value is -125.