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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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The function $f(x)$ is only defined for $x\ge 0$. Also, second derivative has sign error. –  André Nicolas Dec 27 '12 at 4:59
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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? –  Jonas Meyer Dec 27 '12 at 4:59
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up vote 4 down vote accepted

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 $f''(x)=\frac{5}{4}x^{-3/2}$.

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