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A polynomial $\mathrm{P}(\mathrm{x})$ with real coefficients has the property that $\mathrm{P}^{\prime \prime}(\mathrm{x}) \neq 0$ for all $\mathrm{x}$. Suppose $\mathrm{P}(0)=1$ and $\mathrm{P}^{\prime}(0)=-1$. What can you say about $\mathrm{P}(1)$ ?
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The correct answer is:
$\mathrm{P}(1) \leq 0$
$\begin{array}{ll}
\mathrm{P}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \quad \mathrm{P}(0)=1 \\
\mathrm{P}^{\prime}(\mathrm{x})=-\mathrm{e}^{-\mathrm{x}} \quad \mathrm{P}^{\prime}(0)=-1 \quad \mathrm{P}^{\prime \prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \neq 0 \quad \forall \mathrm{x} \in \mathrm{R} \\
\mathrm{P}(1)=\frac{1}{\mathrm{e}}
\end{array}$
$\mathrm{P}(\mathrm{x})=-\mathrm{e}^{\mathrm{x}}+2$
$\mathrm{P}^{\prime}(\mathrm{x})=-\mathrm{e}^{\mathrm{x}}$
$\mathrm{P}^{\prime}(0)=-1$
$\mathrm{P}^{\prime \prime}(\mathrm{x})=-\mathrm{e}^{\mathrm{x}}$
$\mathrm{P}(1)=-\mathrm{e}+2$
$\mathrm{P}(1) \neq 0$
$=-0.7$
\mathrm{P}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \quad \mathrm{P}(0)=1 \\
\mathrm{P}^{\prime}(\mathrm{x})=-\mathrm{e}^{-\mathrm{x}} \quad \mathrm{P}^{\prime}(0)=-1 \quad \mathrm{P}^{\prime \prime}(\mathrm{x})=\mathrm{e}^{\mathrm{x}} \neq 0 \quad \forall \mathrm{x} \in \mathrm{R} \\
\mathrm{P}(1)=\frac{1}{\mathrm{e}}
\end{array}$
$\mathrm{P}(\mathrm{x})=-\mathrm{e}^{\mathrm{x}}+2$
$\mathrm{P}^{\prime}(\mathrm{x})=-\mathrm{e}^{\mathrm{x}}$
$\mathrm{P}^{\prime}(0)=-1$
$\mathrm{P}^{\prime \prime}(\mathrm{x})=-\mathrm{e}^{\mathrm{x}}$
$\mathrm{P}(1)=-\mathrm{e}+2$
$\mathrm{P}(1) \neq 0$
$=-0.7$
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