Let f: R rightarrow (0, infty) and g: R right | vedclass

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(A,D) Given $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$. Since $f^{\prime}(2) = 0$ and $g(2) = 0$,this is a $\frac{0}{0

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$f$ has a local minimum at $x=2$

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(A,D) Given $\lim_{x \rightarrow 2} \frac{f(x) g(x)}{f^{\prime}(x) g^{\prime}(x)} = 1$. Since $f^{\prime}(2) = 0$ and $g(2) = 0$,this is a $\frac{0}{0}$ form. Applying $L$'Hopital's rule:$\lim_{x \rightarrow 2} \frac{f^{\prime}(x) g(x) + f(x) g^{\prime}(x)}{f^{\prime \prime}(x) g^{\prime}(x) + f^{\prime}(x) g^{\prime \prime}(x)} = 1$.Substituting $x=2$ and using $f^{\prime}(2) = g(2) = 0$:$\frac{0 + f(2) g^{\prime}(2)}{f^{\prime \prime}(2) g^{\prime}(2) + 0} = 1 \implies \frac{f(2)}{f^{\prime \prime}(2)} = 1 \implies f(2) = f^{\prime \prime}(2)$.Since $f: R \rightarrow (0, \infty)$,$f(2) > 0$,so $f^{\prime \prime}(2) > 0$. By the second derivative test,$f$ has a local minimum at $x=2$ (Option $A$).Consider $h(x) = f(x) e^{-x^2/2}$. This is not directly helpful,but consider $h(x) = f(x) e^{-x/2}$ or similar. Actually,consider $h(x) = f(x) - f^{\prime \prime}(x)$. Since $f(2) = f^{\prime \prime}(2)$,$h(2) = 0$. Thus,$f(x) - f^{\prime \prime}(x) = 0$ for at least $x=2$ (Option $D$).Thus,both $A$ and $D$ are correct.

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