find the maximum and minimum value of the function f(x)= x-x^3 on the interval [-2,2]
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What is the largest possible product of two negative number whose sum is 1?
check end points
f(-2) = -2+8 = 6
f(2) = 2-8 = -6
now find where derivative is zero
f' = 1 -3x^2
= 0 when x = 1/sqrt3 or x = -1/sqrt3
if x = 1/sqrt 3
f = 1/sqrt3 -1/3sqrt3 =(2/3sqrt3)
if x = -1/sqrt3
f = -1/sqrt3 +1/3sqrt 3
= -(2/3sqrt3)
so the max and min are at the end points max at -2 and min at +2
To find the maximum and minimum values of a function, we need to find the critical points of the function within the given interval and then compare the function values at those points.
1. Finding the critical points of the function f(x) = x - x^3 on the interval [-2, 2]:
To find the critical points, we need to find where the derivative of the function equals zero or is undefined.
First, let's find the derivative of f(x):
f'(x) = 1 - 3x^2
Set f'(x) = 0 and solve for x:
1 - 3x^2 = 0
3x^2 = 1
x^2 = 1/3
x = ±sqrt(1/3)
Since the interval is [-2, 2], we only need to consider the critical points that fall within this range.
The critical points within the interval [-2, 2] are -sqrt(1/3) and sqrt(1/3).
2. Comparing the function values at the critical points:
Evaluate f(x) at the critical points:
f(-sqrt(1/3)) = -sqrt(1/3) - (-sqrt(1/3))^3
f(sqrt(1/3)) = sqrt(1/3) - (sqrt(1/3))^3
Also, evaluate f(x) at the endpoints of the interval:
f(-2) = -2 - (-2)^3
f(2) = 2 - 2^3
Compare the function values:
- Calculate f(-sqrt(1/3)), f(sqrt(1/3)), f(-2), and f(2) using a calculator or by simplifying the expressions.
The maximum value of the function f(x) = x - x^3 on the interval [-2, 2] is the larger value between f(-sqrt(1/3)) and f(2), while the minimum value is the smaller value between f(sqrt(1/3)) and f(-2).
Now, let's solve the second question:
To find the largest possible product of two negative numbers whose sum is 1:
Let's call the two negative numbers x and y, such that x + y = 1.
1. Express one variable in terms of the other:
We can express y = 1 - x.
2. Form the product:
The product of x and y is xy = x(1 - x) = x - x^2.
3. Determine the maximum value:
To find the maximum value, we need to find the vertex of the quadratic equation x - x^2.
The vertex of a quadratic equation in the form ax^2 + bx + c is given by x = -b / (2a).
In this case, a = -1, b = 1, and c = 0. Therefore, the vertex is x = -1 / (2*(-1)) = -1/2.
Substitute this value back into the equation xy = x - x^2:
xy = (-1/2) - (-1/2)^2
Calculate the value of xy using a calculator or by simplifying the expression.
The largest possible product of two negative numbers whose sum is 1 is the value of xy obtained above.