1. **State the problem:** Find all relative maxima and minima of the function $g$ given its derivative $g'(x)$. 2. **Recall the critical points:** Critical points occur where $g'(x) = 0$ or $g'(x)$ is undefined. 3. **Use the first derivative test:** If $g'(x)$ changes from positive to negative at a critical point, $g$ has a relative maximum there. If $g'(x)$ changes from negative to positive, $g$ has a relative minimum. --- ### Problem 4: $g'(x) = (x + 4)e^x$ 4.1. Find critical points by solving $g'(x) = 0$: $$ (x + 4)e^x = 0 $$ Since $e^x > 0$ for all $x$, we have: $$ x + 4 = 0 \implies x = -4 $$ 4.2. Test intervals around $x = -4$: - For $x < -4$, say $x = -5$, $g'(-5) = (-5 + 4)e^{-5} = (-1)(positive) = negative$ - For $x > -4$, say $x = 0$, $g'(0) = (0 + 4)e^0 = 4(1) = positive$ 4.3. Since $g'(x)$ changes from negative to positive at $x = -4$, $g$ has a **relative minimum** at $x = -4$. --- ### Problem 5: $g'(x) = x^2 + 5x + 4$ 5.1. Factor the quadratic: $$ g'(x) = (x + 4)(x + 1) $$ 5.2. Find critical points by solving $g'(x) = 0$: $$ (x + 4)(x + 1) = 0 \implies x = -4, -1 $$ 5.3. Test intervals around $x = -4$ and $x = -1$: - For $x < -4$, say $x = -5$, $g'(-5) = (-5 + 4)(-5 + 1) = (-1)(-4) = positive$ - For $-4 < x < -1$, say $x = -2$, $g'(-2) = (-2 + 4)(-2 + 1) = (2)(-1) = negative$ - For $x > -1$, say $x = 0$, $g'(0) = (0 + 4)(0 + 1) = 4(1) = positive$ 5.4. At $x = -4$, $g'(x)$ changes from positive to negative, so $g$ has a **relative maximum** at $x = -4$. 5.5. At $x = -1$, $g'(x)$ changes from negative to positive, so $g$ has a **relative minimum** at $x = -1$. --- **Final answers:** - Problem 4: Relative minimum at $x = -4$. - Problem 5: Relative maximum at $x = -4$, relative minimum at $x = -1$.