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31 Jul 2008, 10:42
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if x is an integer, which value of x will yield the largest value for 5^-(x-3)^2
x= 3, -3, 9, -9, 0
Note: I saw something similar to this on my own exam..the numbers are made-up of course, the function is madeup also..just trying to test the concept..
x= 3, -3, 9, -9, 0
Note: I saw something similar to this on my own exam..the numbers are made-up of course, the function is madeup also..just trying to test the concept..
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31 Jul 2008, 10:50
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I think the answer would be 3.
regardless of x being greater or less than 3, the - in the front will switch the sign after squaring and make the exponent negative for sure. So we will have \(\frac{1}{5^n}\) where \(n = (x-3)^2\)
If x = 3, then we have -(3-3)^2, or 0^2 = 0, and -0 = 0, and so we then have \(\frac{1}{5^0} = 1/1 = 1\).
Every other value listed for x will give an actual power making 5 * 5 * 5 * 5 or whatever that value becomes. With 5 being in the denominator, the largest fraction will be the one with the smallest denominator.
regardless of x being greater or less than 3, the - in the front will switch the sign after squaring and make the exponent negative for sure. So we will have \(\frac{1}{5^n}\) where \(n = (x-3)^2\)
If x = 3, then we have -(3-3)^2, or 0^2 = 0, and -0 = 0, and so we then have \(\frac{1}{5^0} = 1/1 = 1\).
Every other value listed for x will give an actual power making 5 * 5 * 5 * 5 or whatever that value becomes. With 5 being in the denominator, the largest fraction will be the one with the smallest denominator.
fresinha12
31 Jul 2008, 10:51
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fresinha12
If I'm reading this right 5^-(x-3)^2 = 1/(5^(x-3)^2)
That means you're trying to minimize 5^(x-3)^2
And from that, you're trying to minimize |x-3|
Of the answer choices, 3 would be the best choice.
