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\problemgrouptitle{Antiderivatives}

\item Why are $F_1(x)=\displaystyle\frac{1}{4}x^4$ and $F_2(x)=\displaystyle\frac{1}{4}(x^4+2)$ both antiderivatives of $f(x)=x^3$\,?
\answer{Both $F_1$ and $F_2$ differentiate to $x^3$.  As this problem suggests, any two antiderivatives of a function differ at most by a constant.}

\begin{problemblock}{Find the antiderivative of the given functions.}

\item $\displaystyle f(x)=3x^2-5x+6$
\answer{$\displaystyle F(x)=x^3-\frac{5}{2}x^2+6x+C$}

\item $\displaystyle f(x)=\frac{x^3+4}{x^2}$
\answer{$\displaystyle F(x)=\frac{1}{2}x^2-\frac{4}{x}+C$}

\item $\displaystyle g(t)=\sqrt{t}+\frac{2}{\sqrt{t}}$
\answer{$\displaystyle G(t)=\frac{2}{3}t^{\frac{3}{2}}+4t^{\frac{1}{2}}+C$}

\item $\displaystyle h(x)=\sqrt[3]{x^2}-4x^6+\pi$
\answer{$\displaystyle H(x)=\frac{3}{5}x^{\frac{5}{3}}-\frac{4}{7}x^7+\pi x+C$}

\item $\displaystyle f(\theta)=2\cos\theta-\sin\theta+\sec^2\theta$
\answer{$\displaystyle F(\theta)=2\sin\theta+\cos\theta+\tan\theta+C$}

\end{problemblock}

\begin{problemblock}{Find the function(s) $f$ satisfying the following.}

\item $\displaystyle f''(x)=2x^3-10x+3$
\answer{$\displaystyle f(x)=\frac{1}{10}x^5-\frac{5}{3}x^3+\frac{3}{2}x^2+Cx+D$}

\item $\displaystyle f''(t)=\sqrt{t}$ $\displaystyle +6t,\ \ f(1)=1,\ \ \ f'(1)=2$
\answer{$\displaystyle f(t)=\frac{4}{15}t^{\frac{5}{2}}+t^3-\frac{5}{3}t+\frac{7}{5}$}

\item $\displaystyle f''(\theta )=3\sin \theta +\cos\theta+5,\ \ \ \ f(0)=3,\ \ f'(0)=-1$
\answer{$\displaystyle f(\theta)=-3\sin\theta-\cos\theta+\frac{5}{2}\theta^2+2\theta+4$}

\end{problemblock}

\item Suppose $f$ is a function with $f'''(x)=0$ for all $x$.  Show that $f$ has no points of inflection.
\answer{If $f'''(x)=0$, then $f''(x)=C$ where $C$ is a real constant.  If $C$ is non-zero then $f$ has the same concavity everywhere, while if $C=0$ then $f''(x)=0$ implies $f(x)=Dx+E$ so $f$ is linear and hence has no point of inflection.}
