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$ f(x) = -5 \sqrt{x}$ I want to test the concavity, and I do this $ f'(x) = -\frac{5}{2} x^{-1/2}$ $ f''(x) = -\frac{5}{4} x^{-3/2}$ but if the $x \lt 0$, then $f''(x)$ become complex number, then what should I do? Thx in advance.

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    In what context do you encounter this problem? Are you sure that $f$ is not restricted to the positive real line, and that negative values of $x$ are relevant?2012-12-27

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If this is a typical calculus course, we are only interested in real-valued functions. Thus $f(x)$ is only defined for $x\ge 0$. The first derivative, and higher derivatives, are only defined when $x\gt 0$.

There is a sign issue in your calculation of $f''(x)$ that would affect conclusions about concavity. Note that $f''(x)=\frac{5}{4}x^{-3/2}$.