About extreme values of $\{f(x)-x\}^2$ when $f(x)$ is a cubic function.

Question

• $t \ge 6$, $t \in \mathbb{R}$
• $f(x) = \frac{1}{t}\left( \frac{1}{8}x^3 + \frac{t^2}{8}x+2\right)$
• $\{f(x)-x\}^2$ has an extreme value on $x = k$
• Sum of such $k = g(t)$
• $g(p) = -1$ for some $p \in \mathbb{R}$ $$\int_{6}^{p} g'(t)(8t-t^2)dt = \,?$$

My approach:

I. For condition 3:

1. Let $h(x) = \{f(x)-x\}^2$.
2. Since $h(x)$ has an extreme value on $x=k$, $h'(k) = 0$ and $h''(k) \ne 0$.
3. For $h'(k) = 0$, $h'(k) = 2\{f(k)-k\}(f'(k)-1) =0$, so $f(k)=k$ or $f'(k)=1$.
4. For $h''(k) \ne 0$, $h''(k)=2\left[\{(f'(k)-1)^2 + (f(k)-k)f''(k)\}\right] \ne 0$
5. Combining both 3 and 4,

• If $f(k) = k$, $f'(k) \ne 1$.
• If $f'(k) = 1$, $f''(k)\ne0 \rightarrow k \ne 0$ and $f(k) \ne k$.

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II. Applying $f(x)$:

1. Equation $f(k)=k$ becomes $8t-t^2=k^2 + \frac{16}{k}$.
2. Equation $f'(k)=1$ becomes $8t-t^2=3k^2$.
3. So, step I-5 becomes like:

• If $8t-t^2=k^2 + \frac{16}{k}$, $8t-t^2\ne3k^2$.
• If $8t-t^2=3k^2$, $8t-t^2\ne k^2 + \frac{16}{k}$ and $k \ne 0$.

4. We can draw graphs of $y=3k^2$, $y=k^2 + \frac{16}{k}$, and $y = 8t-t^2$ which will be a constant graph.
5. And we observe points which satisfies step 3.
6. Constant graph($y=8t-t^2$) has a value of less or equal than $12$ since $t \ge 6$ from condition.
7. But the intersect $(2, 12)$(where $t=6$) is the only point where it does not meet condition of step 3, since it is on both functions.
8. So, we can say that $g(6) =(-4) + (-2) = -6$.
9. Below that, all intersects are correct points, so the sum $g(t)$ becomes the $x$-coordinate which satisfies $x^2 + \frac{16}{x} = 8t-t^2$, since sum of $x$-coordinates on $y=3x^2$ becomes $0$ due to symmetry.
10. So we can say that $\left(g(t) \right)^2 + \frac{16}{g(t)} = 8t - t^2$.

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III. Getting the answer. $$\int_{6}^{p} g'(t)(8t-t^2)dt = \int_{6}^{p} g'(t)\left(\left(g(t) \right)^2 + \frac{16}{g(t)}\right)dt $$

Let $g(t) = s$, $$\int_{g(6)}^{g(p)} \left(s^2 + \frac{16}{s}\right)ds$$

Since $g(p) = -1$ from condition and $g(6) = -6$ from what we've got, $$\int_{-6}^{-1} \left(s^2 + \frac{16}{s}\right)ds = \left[ \frac{1}{3}s^3 + 16\ln{\left| s\right|} \right]_{-6}^{-1} = \frac{215}{8}-\ln{6}$$

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And it was wrong. The correct answer was $21 - 32\ln2$, and it seems like the value $g(6)$ was slightly off.

I tried to explain my process as specific as possible, because I want to know if I am making a inefficient approach or I use too vague logic. And, the reason of my wrong answer. Any comment would be so helpful to me right now.

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+edit: By If a function $f(x)$ has an extreme value on $x=k$, $f''(k) \ne 0$? , step I-2 is a completely wrong logic. But $h(x)$ is a polynomial function, so checking $h'(x)=0$ is still a valid approach(not that it approves it is an extrema).

By manually checking if the point $k=2$ is also an extrema, $k^2 + \frac{16}{k}$ is a $(+)$, and $3k^2$ goes from $(-)$ to $(+)$, it becomes an extrema.

Thus, $g(6)=-4$, and we can get the correct answer.