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The derivative of a function f is given by $f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+2}$ , for $x \in \mathbb{R}$ .

Hence, justify that the value of x found in part (b)(i) corresponds to a local minimum point on the graph of f.

It is given that $f(2)=3+\ln 10$ .

substituting x=-1 into $f^{\prime \prime}(x)$ , clearly leading to positive numerator A1

\begin{aligned} & f^{\prime \prime}(-1)=\left(\frac{-2+4}{1}\right)(=2) \\ & f^{\prime \prime}(-1)>0 \end{aligned}

therefore this is a local minimum point AG

Note: Do not award AOR1.

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