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\problemgrouptitle{Inflection Points}
\begin{problemblock}{Show that the following functions have no inflection points:}
\item $f(x)=2x^4$
\answer{Notice that $f''(x)=24x^2$ and $24x^2$ is positive for all values of $x$ except 0.  The only potential inflection point would therefore occur at $x=0$ but the concavity is positive on both sides of $x=0$ and hence does not change at that value.
}

\item $\displaystyle f(x)=\frac{1}{x}$
\answer{Notice that $\displaystyle f''(x)=\frac{2}{x^3}$ which is negative for $x<0$ and positive for $x>0$.  Therefore the concavity does, in fact, change at $x=0$.  However the function is not defined at $0$ so there is no point on the curve there (it is a vertical asymptote) and hence no inflection point exists.}

\end{problemblock}

\item Find the \relative extrema (and their locations), the intervals of concavity, and the inflection points of the function $f(x)=x^5-15x^3+1$\,.
\answer{Notice that $f'(x)=5x^2(x-3)(x+3)$, $f''(x)=10x\left(\sqrt{2}x-3\right)\left(\sqrt{2}x+3\right)$.  \Relative maximum: $f(-3)=163$; \Relative minimum: $f(3)=-161$;  Inflection points: $\displaystyle\left(-\frac{3}{\sqrt{2}},-\frac{243}{{2}^{\frac{5}{2}}}+\frac{405}{{2}^{\frac{3}{2}}}+1\right)$,  $(0,1)$, $\displaystyle\left(\frac{3}{\sqrt{2}},\frac{243}{{2}^{\frac{5}{2}}}-\frac{405}{{2}^{\frac{3}{2}}}+1\right)$; Concave upward on: $\displaystyle\left(-3/\sqrt{2},0\right)\cup\left(3/\sqrt{2},\infty\right)$; Concave downward on: $\displaystyle\left(-\infty,-3/\sqrt{2}\right)\cup\left(0,3/\sqrt{2}\right)$;
}
