Which one of the following functions is increasing on the set (0,10]? Group of answer choices g(x) = 7 - 3x g(x) = x² - 30x g(x) = -1 / x g(x) = sin(x)
Question
Which one of the following functions is increasing on the set (0,10]? Group of answer choices
g(x) = 7 - 3x
g(x) = x² - 30x
g(x) = -1 / x
g(x) = sin(x)
Solution
To determine which function is increasing on the set (0,10], we need to look at the derivative of each function.
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g(x) = 7 - 3x The derivative of this function is g'(x) = -3. Since the derivative is negative, this function is decreasing, not increasing.
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g(x) = x² - 30x The derivative of this function is g'(x) = 2x - 30. To find where this function is increasing, we set the derivative equal to zero and solve for x: 2x - 30 = 0, x = 15. This means the function is increasing for x > 15, which does not include the interval (0,10]. So, this function is not increasing on the interval (0,10].
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g(x) = -1 / x The derivative of this function is g'(x) = 1 / x². Since the derivative is positive for x > 0, this function is increasing on the interval (0,10].
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g(x) = sin(x) The derivative of this function is g'(x) = cos(x). The cosine function oscillates between -1 and 1, so the sine function is not always increasing.
Therefore, the function that is increasing on the set (0,10] is g(x) = -1 / x.
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