Let $a$ be a positive real number. We are to investigate local extrema of the function $$f(x) = x^2 - 5 + 4a\log(2x + a + 8) \quad \left(-\frac{a}{2} - 4 < x < -2\right).$$ (1) When we differentiate the function $f(x)$ with respect to $x$, we obtain $$f'(x) = \frac{\mathbf{A}(\mathbf{B}\, x + a)(x + \mathbf{C})}{\mathbf{D}\, x + a + \mathbf{E}}.$$ (2) Since a condition of $a$ is that $a > 0$ and the domain of $f(x)$ is $-\frac{a}{2} - 4 < x < -2$, the range of values of $a$ such that $f(x)$ has both a local maximum and a local minimum is $$\mathbf{F} < a < \mathbf{G}.$$ In such a case, the sum of the local maximum and the local minimum is $$\frac{a^2}{\mathbf{H}} + \mathbf{I} + \mathbf{IJ}\, a\log\mathbf{K}\, a.$$
Let $a$ be a positive real number. We are to investigate local extrema of the function
$$f(x) = x^2 - 5 + 4a\log(2x + a + 8) \quad \left(-\frac{a}{2} - 4 < x < -2\right).$$
(1) When we differentiate the function $f(x)$ with respect to $x$, we obtain
$$f'(x) = \frac{\mathbf{A}(\mathbf{B}\, x + a)(x + \mathbf{C})}{\mathbf{D}\, x + a + \mathbf{E}}.$$
(2) Since a condition of $a$ is that $a > 0$ and the domain of $f(x)$ is $-\frac{a}{2} - 4 < x < -2$, the range of values of $a$ such that $f(x)$ has both a local maximum and a local minimum is
$$\mathbf{F} < a < \mathbf{G}.$$
In such a case, the sum of the local maximum and the local minimum is
$$\frac{a^2}{\mathbf{H}} + \mathbf{I} + \mathbf{IJ}\, a\log\mathbf{K}\, a.$$