Q.(a) Find the stationary points and stationary values for the function : $f(x) = 2x^{3} + 9x^{2} + 12x + 1$.

(a) $f'(x)=6(x+1)(x+2)$; stationary points $x=-1,-2$ with values $-4,-3$. (b) Linear fit: $y=6x+3000$.

Part (a) — Stationary points and values of $f(x)=2x^{3}+9x^{2}+12x+1$.

Step 1 — Differentiate and set to zero.

$f'(x)=6x^{2}+18x+12=6(x^{2}+3x+2)=6(x+1)(x+2)=0.$

$\Rightarrow x=-1\ \text{or}\ x=-2.$

Step 2 — Stationary values.

$f(-1)=2(-1)+9(1)+12(-1)+1=-2+9-12+1=-4.$

$f(-2)=2(-8)+9(4)+12(-2)+1=-16+36-24+1=-3.$

Step 3 — Nature (using $f''(x)=12x+18$).

$f''(-1)=6>0$ (relative minimum, value $-4$); $f''(-2)=-6<0$ (relative maximum, value $-3$).

Part (b) — Linear cost relationship.

Let $y=ax+b$ pass through $(x,y)=(500,6000)$ and $(1000,9000)$.

…