Determine where the absolute extrema of f(x)=-3(x^2)+7x on the interval [1,3] occur.
1. The absolute maximum occurs at =
2. The absolute minimum occurs at =
Set the derivative f'(x) = 0 and solve for x.
-6x + 7 = 0
x = 7/6
Since the second derivative f"(x) = -6, the extremum at x = 7/6 is a maximum. The minimum for the interval [1,3] will occur at x = 3, since the f(x) just keeps getting lower for higher values of x when x > 7/6.
To find the absolute extrema of the function f(x)=-3x^2+7x on the interval [1,3], we need to first find the critical points and the endpoints of the interval.
1. Find the critical points:
A critical point of a function occurs where its derivative is either zero or undefined.
To find the critical points of f(x), we need to find f'(x) and set it to zero.
f(x) = -3x^2 + 7x
f'(x) = -6x + 7
Setting f'(x) = 0:
-6x + 7 = 0
-6x = -7
x = 7/6
So, the critical point is x = 7/6.
2. Evaluate the function at the critical point and the endpoints:
Evaluate f(x) at x = 1, x = 3, and x = 7/6 to find the function values at these points.
f(1) = -3(1)^2 + 7(1) = -3 + 7 = 4
f(3) = -3(3)^2 + 7(3) = -27 + 21 = -6
f(7/6) = -3(7/6)^2 + 7(7/6) ≈ 5.5
So, the function values at the critical point and endpoints are:
f(1) = 4
f(3) = -6
f(7/6) ≈ 5.5
3. Determine the absolute extrema:
The absolute maximum of a function on a closed interval is the highest point on the function, and the absolute minimum is the lowest point.
Comparing the function values, we see that the absolute maximum occurs at x = 1 with a value of f(1) = 4.
The absolute minimum occurs at x = 3 with a value of f(3) = -6.
Therefore:
1. The absolute maximum of f(x) = -3x^2 + 7x on the interval [1,3] occurs at x = 1 with a value of 4.
2. The absolute minimum of f(x) = -3x^2 + 7x on the interval [1,3] occurs at x = 3 with a value of -6.