Near $x = 2$ , how does $f(x) = \log x - x$ behave?

Strictly increasingCorrect answerYour answerNeither: the derivative is 0 thereCorrect answerYour answerConstantCorrect answerYour answerStrictly decreasingCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$f'(x) = \frac{1}{x} - 1$ , so $f'(2) = \frac{1}{2} - 1 = -\frac{1}{2}$ .
A negative derivative means $f$ is falling there.

Worth noting

$\log x$ itself rises, but here it rises more slowly than $x$ , so the difference falls.

Concept: Increasing or decreasing at a point
Typical timeTypical: 60 s
Your timeYours: —

The rule this testsThe derivative test

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Ten questions from Increasing & Decreasing Functions, at your own pace.

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Near x=2x = 2x=2, how does f(x)=log⁡x−xf(x) = \log x - xf(x)=logx−x behave?

Explanation

Near $x = 2$ , how does $f(x) = \log x - x$ behave?