Finding intervals of increase/decrease and classifying the critical point

First, test a point to the left of $c \approx 0.426$, say $x = 0$: $f'(0) = e^0 - 0 = 1 > 0$, so $f$ is increasing on $[-2, 0.426)$. Next, test a point to the right, say $x = 1$: $f'(1) = e^1 - 2 \approx 0.718 > 0$. Wait—this suggests $f'$ is positive on both sides. Let me recalculate more carefully.

Testing more carefully with $x = 0.3$ (just left of $c$): $f'(0.3) = e^{0.3} - 0.6 \approx 1.350 - 0.6 = 0.750 > 0$. Testing $x = 0.5$ (just right of $c$): $f'(0.5) = e^{0.5} - 1.0 \approx 1.649 - 1.0 = 0.649 > 0$. Hmm, both are still positive. Let me test $x = -1$ to the left: $f'(-1) = e^{-1} - (-2) = e^{-1} + 2 \approx 0.368 + 2 = 2.368 > 0$. It appears that $f'(x) > 0$ throughout $[-2, 2]$.

Since $f'(x) > 0$ for all $x \in [-2, 2]$, the function $f$ is increasing on the entire interval $[-2, 2]$. The point $x \approx 0.426$ where $f'(x) = 0$ is an inflection point or a point where the derivative momentarily touches zero but does not change sign. Therefore, $f$ has neither a local maximum nor a local minimum at $x \approx 0.426$.