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If $f(x)=b e^{a x}+a e^{b x}$, then $f^{\prime \prime}(0)=$
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Verified Answer
The correct answer is:
$a b(a+b)$
We have,
$$
\begin{aligned}
& f(x) &=b e^{a x}+a e^{b x} \\
\Rightarrow & f^{\prime}(x) &=a b e^{a x}+a b e^{b x} \\
\Rightarrow & & f^{\prime \prime}(x) &=a^{2} b e^{a x}+a b^{2} e^{b x} \\
\therefore & & f^{\prime \prime}(0) &=a^{2} b+a b^{2}=a b(a+b)
\end{aligned}
$$
$$
\begin{aligned}
& f(x) &=b e^{a x}+a e^{b x} \\
\Rightarrow & f^{\prime}(x) &=a b e^{a x}+a b e^{b x} \\
\Rightarrow & & f^{\prime \prime}(x) &=a^{2} b e^{a x}+a b^{2} e^{b x} \\
\therefore & & f^{\prime \prime}(0) &=a^{2} b+a b^{2}=a b(a+b)
\end{aligned}
$$
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