Question
Let f: [3, β) β R be defined by f(x) = xΒ² β 6x +
11. The range of f is:Solution
We are given the function: f(x) = xΒ² β 6x + 11, with domain x β [3, β) We are to find the range of f(x). Given: f(x) = xΒ² β 6x + 11 = (x β 3)Β² + 2 This is a perfect square shifted: It's a parabola opening upward, with minimum value at x = 3. At x = 3: f(3) = (3 β 3)Β² + 2 = 0 + 2 = 2 As x increases, (x β 3)Β² increases, so f(x) increases. Hence: Minimum value = 2 Maximum = β (since x β β β f(x) β β) So: Range = [2, β)
The value of 0∫π cos3x dx is –
44.98% of 399.98 = ? x 19.98 - (39.98% of 2499.98)
Maximize Z = 3x + 5y
Subject to:
x β y β₯ 2
x + y β₯ 4
x, y β₯ 0
What can be concluded about the feasible region?
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