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17 Jan 2022, 03:15
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D
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Difficulty:
75%
(hard)
Question Stats:
51% (02:01) correct
49%
(01:51)
wrong
based on 258
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Let f(x) = min{x^2 - 2, 1 - 2x}, for any positive number x. What is the maximum value of f(x) ?
A. 2
B. 1
C. 0
D. -1
E. -2
Are You Up For the Challenge: 700 Level Questions
A. 2
B. 1
C. 0
D. -1
E. -2
Are You Up For the Challenge: 700 Level Questions
31 Jan 2022, 18:08
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Bunuel
Breaking Down the Info:
This function finds the minimum between two values. The left value increases with x, while the right value decreases with x. If we want the maximum value of this function, we should find the equilibrium point.
\(x^2 - 2 = 1 - 2x\) and \(x^2 + 2x - 3 = 0\). Then \((x + 3)(x - 1) = 0\), so \(x = -3\) or \(1\).
We are not allowed to take negative x however, so we can start with x = 1 and find f(1) = -1. We can try f(2) = min(2, -3) and f(0) = min(-2, ?) to see the function only decreases from here. Thus f(1) = -1 is the maximum value.
Answer: D
General Discussion
12 Aug 2025, 16:21
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