 Solution

For negative $x$, the mean value theorem gives $f^{\prime }(x)=3+x{f}^{\prime \prime }({\xi }_x)$. In particular,Since $f^{\prime }$ is continuous and strictly increasing, it vanishes at exactly one point $x_m\in (-2,-1)$. The positive second derivative makes $f$ a strictly convex function on the entire real line. Its derivative changes from negative to positive at $x_m$, so $x_m$ is the unique global minimum. One may also use $f(x)\ge x^2+3x-5\to \infty$ to verify coercivity.

In the endpoint-bracketing form of golden-section search, begin with $[-2,-1]$, a width-one interval known to contain this unique minimum, and place the two standard interior probes. The attainable position accuracy is different from that for a simple root. Around $x_m$, Taylor's theorem givesHere the minimum value is nonzero and of order one: the lower quadratic bound gives $f(x_m)\ge -29/4$, while $f(x_m)\le f(-1)<-13/2$. Ordinary relative round-off error in these function values therefore hides changes of order $ϵ$, once $h^2$ has that size. The function-value resolution near a nondegenerate minimum is consequently of orderEach golden-section search contraction multiplies the bracket width by $\rho =(\sqrt{5}-1)/2$. Using $2^{-20}$ as the nominal target width givesUnknown evaluation-error constants can change the actual stopping point; $29$ is the standard scale-based count. Requiring $2^{-40}$ position accuracy from these function-value comparisons would disregard the quadratic flatness of the minimum. The same precision effect is documented for one-dimensional numerical minimization.^[ref]