Discussion
The function\(f(x) = e^{x} - x\) is increasing in the interval
Answer: E
f is increasing on \((0, \infty)\).
A differentiable function increases where its derivative is positive.
\[ f'(x) = e^{x} - 1 \gt 0 \iff e^{x} \gt 1 \iff x \gt 0 \]
So f increases on the whole of \((0, \infty)\) and decreases on \((-\infty, 0)\), with its minimum value 1 at x = 0. \((-\infty, 0)\) and \((-1, 0)\) are where it decreases, and \((-1, 1)\) straddles the minimum. \((0, 4)\) lies inside the interval of increase, so f increases there too, but it is only part of it; the question asks for the interval in which f increases, which is all of \((0, \infty)\).
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