A calculus problem by Aziz Alasha

If $f(x) = x^2 + 2^x + 2$ , then $f'(x) = 2x + 2^x \cdot ln(2) = 0 \Rightarrow x + 2^{x-1} \cdot ln(2) = 0 \Rightarrow 2^x \cdot ln(\sqrt{2}) = -x.$ This exponential curve and line must intersect in the second quadrant of the $xy-$ plane. At $x = -1$ we obtain $2^{-1} \cdot ln(\sqrt{2}) \approx 0.1733$ for the LHS and $1$ for the RHS, which implies the critical point must lie within $x \in (-1,0).$ After some trial-&-error I find the critical value to be $x \approx -0.2845$ , and $f''(x) = 2 + 2^{x} \cdot (ln(2))^2 > 0$ for all real $x$ . So the minimum value of $f(x)$ computes to $(-0.2845)^2 + 2^{-0.2845} + 2 \approx 2.904.$