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Let M and m be respectively the absolute maximum and the absolute minimum values of the function, $$f(x) = 2x^3 - 9x^2 + 12x + 5$$ in the interval [0, 3]. Then M - m is equal to:
We have to find the absolute (global) maximum and minimum values of the cubic function
$$f(x)=2x^{3}-9x^{2}+12x+5$$
on the closed interval $$[0,3]$$. For a continuous function on a closed interval, the Extreme‐Value Theorem tells us that absolute maxima and minima occur either at the critical points (where the derivative is zero or undefined) that lie inside the interval, or at the end points of the interval.
First, we compute the derivative. Using the power rule $$\dfrac{d}{dx}(x^{n})=nx^{\,n-1},$$ we obtain
$$f'(x)=\dfrac{d}{dx}\bigl(2x^{3}-9x^{2}+12x+5\bigr) =2\cdot3x^{2}-9\cdot2x+12 =6x^{2}-18x+12.$$
Now we factor the derivative to locate its zeros:
$$f'(x)=6x^{2}-18x+12 =6(x^{2}-3x+2) =6\bigl(x^{2}-3x+2\bigr).$$
The quadratic inside can itself be factored:
$$x^{2}-3x+2=(x-1)(x-2).$$
So
$$f'(x)=6(x-1)(x-2).$$
A critical point occurs wherever $$f'(x)=0$$, i.e.
$$6(x-1)(x-2)=0\;\Longrightarrow\;x=1\quad\text{or}\quad x=2.$$ Both $$x=1$$ and $$x=2$$ lie inside the interval $$[0,3]$$, so they are potential places for absolute extrema.
Next, we evaluate the original function at all candidates: the two critical points and the two end points.
• At the left end point $$x=0$$:
$$f(0)=2(0)^{3}-9(0)^{2}+12(0)+5=0-0+0+5=5.$$
• At the critical point $$x=1$$:
$$f(1)=2(1)^{3}-9(1)^{2}+12(1)+5 =2(1)-9(1)+12(1)+5 =2-9+12+5 =10.$$
• At the critical point $$x=2$$:
$$f(2)=2(2)^{3}-9(2)^{2}+12(2)+5 =2(8)-9(4)+24+5 =16-36+24+5 =9.$$
• At the right end point $$x=3$$:
$$f(3)=2(3)^{3}-9(3)^{2}+12(3)+5 =2(27)-9(9)+36+5 =54-81+36+5 =14.$$
We now compare these four values:
$$f(0)=5,\quad f(1)=10,\quad f(2)=9,\quad f(3)=14.$$
The largest of these is $$14$$, so the absolute maximum is
$$M=14\quad\text{(occurs at }x=3).$$
The smallest of these is $$5$$, so the absolute minimum is
$$m=5\quad\text{(occurs at }x=0).$$
Therefore
$$M-m=14-5=9.$$
Hence, the correct answer is Option A.
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