There is a useful approximation to the certainty equivalent that is easy to derive. A second-order expansion near $\bar{x}=\mathrm{E}(x)$ gives

\[U(x) \approx U(\bar{x})+U^{\prime}(\bar{x})(x-\bar{x})+\frac{1}{2} U^{\prime \prime}(\bar{x})(x-\bar{x})^{2} .\]

Hence,

\[\mathrm{E}[U(x)] \approx U(\bar{x})+\frac{1}{2} U^{\prime \prime}(\bar{x}) \operatorname{var}(x) .\]

On the other hand, if we let $c$ denote the certainty equivalent and assume it is close to $\bar{x}$, we can use the first-order expansion

\[U(c) \approx U(\bar{x})+U^{\prime}(\bar{x})(c-\bar{x}) .\]

Using these approximations, show that

\[c \approx \bar{x}+\frac{U^{\prime \prime}(\bar{x})}{U^{\prime}(\bar{x})} \operatorname{var}(x)\]