- For the real number $t$ that $t \ge 6 $, let $f(x) = \frac{1}{t} \left( \frac{1}{8}x^3 + \frac{t^2}{8}x + 2 \right)$.
- Let the sum of all real numbers $k$ satisfying the following condition be $g(t)$ : function $\{f(x)-x\}^2$ has an extreme value at $x=k$.
- For the real number $p$ where $g(p) = -1$, $$\int_{6}^{p}g'(t)(8t-t^2)dt = ?$$
My attempt:
- Since the function $h(x) = \{f(x)-x\}^2$ has an extreme value at $x=k$, by differentiation, $h'(k) = 2\{ f(k)-k \}\left( f'(k)-1 \right) =0$.
- So, $f(k) = k$ or $f'(k) = 1$.
- $h''(k) \ne 0$ because it cannot be an inflection point.
- Applying it,
- if $f(k) = k$, $f'(k) \ne 1$.
- if $f'(k) = 1$, $f(k) \ne k$.
And I'm stuck. It seems like I have to use the equation of $f(x)$, but it gives me tons of equations with messy variables. A little hint would be really helpful.
(+ Additional approach) :
Applying step 4. to $f(x)$, we can get $t(8-t) = k^2 + \frac{16}{k}$ ($k\ne 2$ when $t=6$).
By multiplying $k$ to each side, $k^3 - t(8-t)k + 16 = 0$.
This is a cubic equation with respect to $k$, so I thought I could use $\alpha + \beta + \gamma = -\frac{b}{a}$ to get $g(t)$. But as you can see here, it becomes $0$ since it also counts imaginary roots.
Instead of trying to calculate the second derivative of $h(x)$ directly, consider the polynomial $p(x) = f(x) - x$. How many roots does it have, and how many stationary points does it have? (You may need to consider two or three different cases.)
Now, since $h(x) = p(x)^2$, its stationary points can only happen when $p(x) = 0$ or $p'(x) = 0$. Of course $p(x)$ is a cubic so solving $p(x) = 0$ is probably messy, but if its roots are $\alpha, \beta, \gamma$ do you know something about $\alpha + \beta + \gamma$?
Between the two of those, I think you have enough information to figure some useful things out about $g(t)$.