Let f(x) = (3+2x-x^2)^2 be defined for the interval [-2,3]. If M is the y-coordinate of the absolute maximum and m is the y-coordinate of the absolute minimum of f(x) on the interval, what is the value of (M+m)?
f(x) = (3+2x-x^2)^2
= (3 - x)^2 (1 + x)^2
so there are double roots at x = -1 and x = 3
https://www.wolframalpha.com/input/?i=plot+f%28x%29+%3D+%283%2B2x-x%5E2%29%5E2
f(-2) = 5^2(-1)^2 = 25
f(3) = 0
my graph shows a local max at x = appr 1, but is is lower than (-2,25)
so M = 25 and m = 0
continue
Well, calculating the value of (M+m) is as important as deciding whether to wear a t-shirt with a clown face on it to a job interview. It's a fine balance between standing out and being completely inappropriate.
But fear not! We can solve this math problem without resorting to clownish tactics. So, let's find the absolute maximum and minimum of f(x) on the interval [-2, 3].
First, let's find the derivative of f(x) by applying some mathematical hocus-pocus and a pinch of laughter:
f'(x) = 2(3+2x-x^2)(2-2x)
Now, let's set f'(x) equal to zero and find the critical points (where the derivative is equal to zero or undefined):
2(3+2x-x^2)(2-2x) = 0
Simplifying this equation, we find two critical points: x = -1 and x = 2.
Next, let's evaluate f(x) at these critical points along with the endpoints of the interval:
f(-2) = (3+2(-2)-(-2)^2)^2 = (3-4-4)^2 = 9
f(-1) = (3+2(-1)-(-1)^2)^2 = (3-2-1)^2 = 0
f(2) = (3+2(2)-(2)^2)^2 = (3+4-4)^2 = 9
f(3) = (3+2(3)-(3)^2)^2 = (3+6-9)^2 = 0
Now, we compare these values to find the absolute maximum and minimum:
Absolute maximum: M = 9 (achieved at x = -2 and x = 2)
Absolute minimum: m = 0 (achieved at x = -1 and x = 3)
Finally, we can calculate the value of (M+m):
M + m = 9 + 0 = 9
So, the value of (M+m) is 9.
To find the absolute maximum and minimum of f(x) on the given interval [-2,3], we need to find the critical points and endpoints of the interval where the maximum and minimum values can occur.
1. First, we need to find the critical points by taking the derivative of f(x) with respect to x and solving for when the derivative equals zero.
f(x) = (3 + 2x - x^2)^2
Taking the derivative:
f'(x) = 2(3 + 2x - x^2)(2) = 4(3 + 2x - x^2)
Setting f'(x) = 0:
4(3 + 2x - x^2) = 0
Now, solve for x:
3 + 2x - x^2 = 0
This is a quadratic equation which can be factored:
(x - 3)(x + 1) = 0
So, x = 3 or x = -1 are the critical points.
2. Next, we need to evaluate the function f(x) at the critical points and the interval endpoints to find the maximum and minimum values.
f(-2) = (3 + 2(-2) - (-2)^2)^2 = (3 - 4 - 4)^2 = (-5)^2 = 25
f(-1) = (3 + 2(-1) - (-1)^2)^2 = (3 - 2 - 1)^2 = 0^2 = 0
f(3) = (3 + 2(3) - (3)^2)^2 = (3 + 6 - 9)^2 = 0^2 = 0
So, the critical points and the endpoints of the interval are (-2, 25), (-1, 0), and (3, 0).
3. Finally, we compare the y-coordinates of the critical points and endpoints to determine the absolute maximum and minimum values.
The absolute maximum value occurs at (-2, 25) since it has the highest y-coordinate among all the points.
The absolute minimum value occurs at (-1, 0) and (3, 0) since each has the lowest y-coordinate among all the points.
Therefore, the value of (M+m) = 25 + 0 = 25.
To find the absolute maximum and minimum of f(x) on the interval [-2, 3], you need to follow these steps:
1. Find the critical points of f(x): These are the values of x where the derivative of f(x) is equal to zero or does not exist.
- Calculate the derivative of f(x): f'(x) = 2(3 + 2x - x^2) * (2 - 2x)
- Set f'(x) = 0 and solve for x: 2(3 + 2x - x^2)(2 - 2x) = 0
2. Determine the values of f(x) at the critical points and the endpoints of the interval.
- Evaluate f(x) at x = -2, 3, and the solutions x obtained in step 1.
3. Compare the values obtained in step 2 to find the absolute maximum (M) and the absolute minimum (m) of f(x) on the interval.
Now let's go through the steps in detail:
1. Find the critical points of f(x):
- Calculate the derivative of f(x): f'(x) = 2(3 + 2x - x^2) * (2 - 2x)
- Simplify the expression of f'(x):
f'(x) = 2(6 - 2x - 3x^2 + 4x - 2x^2 - x^2)
f'(x) = 2(-4x^2 + 2x + 6)
f'(x) = -8x^2 + 4x + 12
- Set f'(x) = 0 and solve for x:
-8x^2 + 4x + 12 = 0
To solve this quadratic equation, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -8, b = 4, and c = 12:
x = (-4 ± √(4^2 - 4(-8)(12))) / (2(-8))
x = (-4 ± √(16 + 384)) / -16
x = (-4 ± √400) / -16
Therefore, the two solutions for x are:
x = (-4 + 20)/-16 = -16/16 = -1
x = (-4 - 20)/-16 = -24/-16 = 3/2 = 1.5
2. Determine the values of f(x) at the critical points and the endpoints of the interval:
- Evaluate f(x) at x = -2, x = 3, and x = -1:
- f(-2) = (3 + 2(-2) - (-2)^2)^2 = (3 - 4 - 4)^2 = (-5)^2 = 25
- f(3) = (3 + 2(3) - (3)^2)^2 = (3 + 6 - 9)^2 = 0
- f(-1) = (3 + 2(-1) - (-1)^2)^2 = (3 - 2 - 1)^2 = (0)^2 = 0
3. Compare the values obtained in step 2 to find the absolute maximum (M) and the absolute minimum (m) of f(x) on the interval:
- The maximum value is M = 25, which occurs at x = -2.
- The minimum value is m = 0, which occurs at x = 3 and x = -1.
Finally, to find (M + m):
M + m = 25 + 0 = 25.
Therefore, the value of (M + m) is 25.