Determine the maximum and minimum of each function on the given interval.
a) ���� = 2x^3�� − 9x^2�� ,−2 ≤ x� ≤ 4
b) ���� = 12x� − x^3�� , � x∈ �[−3,5]
I will do the first one, you do the second in the same way
a) y = 2x^3 - 9x^2 , -2 ≤ x ≤ 4
dy/dx = 6x^2 - 18x
= 0 for a max/min
6x(x-3) = 0
x = 0 or x = 3, both are within our given domain
if x = 0 , y = 0
if x = 3 , y = 56 - 81 = -27
so the max is 0 and the min is -27
should have checked the end points
if x = -2 , y = -16 - 36 = -52
if x = 4 , y = -16
revised answer:
the max is 0 and the min is -52 within the given domain.
To find the maximum and minimum of a function on a given interval, you need to follow these steps:
Step 1: Find the critical points of the function within the given interval by taking the derivative and setting it equal to zero.
Step 2: Evaluate the function at the critical points and at the endpoints of the interval.
Step 3: Compare the values obtained in step 2 to determine the maximum and minimum.
Let's find the maximum and minimum of each function on the given intervals.
a) The function is f(x) = 2x^3 - 9x^2, and the interval is -2 ≤ x ≤ 4.
Step 1: Find the derivative of f(x):
f'(x) = 6x^2 - 18x
Now, set f'(x) = 0 to find the critical points:
6x^2 - 18x = 0
6x(x - 3) = 0
x = 0 or x = 3
Step 2: Evaluate the function at the critical points and the endpoints of the interval:
f(-2) = 2(-2)^3 - 9(-2)^2 = 32 - 36 = -4
f(0) = 2(0)^3 - 9(0)^2 = 0
f(3) = 2(3)^3 - 9(3)^2 = 54 - 81 = -27
f(4) = 2(4)^3 - 9(4)^2 = 128 - 144 = -16
Step 3: Compare the values obtained in step 2:
The maximum value is 0, and the minimum value is -27.
b) The function is g(x) = 12x - x^3, and the interval is -3 ≤ x ≤ 5.
Step 1: Find the derivative of g(x):
g'(x) = 12 - 3x^2
Now, set g'(x) = 0 to find the critical points:
12 - 3x^2 = 0
3x^2 = 12
x^2 = 4
x = -2 or x = 2
Step 2: Evaluate the function at the critical points and the endpoints of the interval:
g(-3) = 12(-3) - (-3)^3 = -36 + 27 = -9
g(-2) = 12(-2) - (-2)^3 = -24 + 8 = -16
g(2) = 12(2) - (2)^3 = 24 - 8 = 16
g(5) = 12(5) - (5)^3 = 60 - 125 = -65
Step 3: Compare the values obtained in step 2:
The maximum value is 16, and the minimum value is -65.
Therefore, the maximum and minimum of the given functions on their respective intervals are:
a) Maximum: 0, Minimum: -27
b) Maximum: 16, Minimum: -65