Let f(x) = a(x - 5)^3 + b(x - 5) + 7. If the solution of 1 <= f(x) <= 13 is k - 9 <= x <= 3k + 7, find k.

Step-by-step solution

1. 1

   Strip the function down to its odd core. Write $t=x-5$ and $g(t)=at^3+bt$. Then

   $f(x)=g(x-5)+7$

   and $g$ is an odd function, since $g(-t)=-at^3-bt=-g(t)$. Every cube-plus-linear expression behaves this way — no even powers, so nothing breaks the symmetry.

2. 2

   Turn oddness into point symmetry of $f$. For any $t$,

   $f(5+t)+f(5-t)=\left[g(t)+7\right]+\left[g(-t)+7\right]=g(t)-g(t)+14=14$

   so the graph of $f$ is symmetric about the point $(5,7)$: whatever happens $t$ units right of $x=5$ is mirrored, upside down, $t$ units to the left.

3. 3

   Observe that the bounds are symmetric about 7 too. The constraint is

   $1\le f(x)\le 13\iff |f(x)-7|\le 6$

   since $1=7-6$ and $13=7+6$. A condition symmetric about the centre value, imposed on a function symmetric about the centre point, must have a solution set symmetric about $x=5$.

4. 4

   Force the given interval to be centred at 5. The midpoint of $\left[k-9,\;3k+7\right]$ is

   $\frac{(k-9)+(3k+7)}{2}=\frac{4k-2}{2}=2k-1$

   Setting this equal to $5$:

   $2k-1=5\;\Longrightarrow\;k=3$

5. 5

   Check that $k=3$ gives a genuine interval. With $k=3$ the endpoints are $k-9=-6$ and $3k+7=16$, so the solution set is $[-6,16]$ — a valid non-empty interval whose midpoint is $\tfrac{-6+16}{2}=5\;\checkmark$, matching the centre of symmetry. Its half-width $11$ then constrains $a$ and $b$ through $f(16)=13$, but $k$ itself is fixed by the symmetry alone.