Let $$f:$$ $$\mathbb{R}$$ $$\longrightarrow\ $$ $$\mathbb{R}$$ be $$f(x)=x^3-6x^2+9x+1$$. Let $$M$$ and $$m$$ denote the local maximum value and the local minimum value of $$f$$ respectively. What is the value of $$Mm$$?

To find the local maximum value $$M$$ and the local minimum value $$m$$ of the function $$f(x) = x^3 - 6x^2 + 9x + 1$$, let us find its critical points using the first derivative test.

Differentiate $$f(x)$$ with respect to $$x$$:

$$f'(x) = 3x^2 - 12x + 9$$

Set $$f'(x) = 0$$ to find the critical points:

$$3(x^2 - 4x + 3) = 0$$

$$3(x - 1)(x - 3) = 0$$

This gives two critical points:

$$x = 1 \quad \text{and} \quad x = 3$$

Now, examine the second derivative to determine the nature of these critical points:

$$f''(x) = 6x - 12$$

• At $$x = 1$$:
  $$f''(1) = 6(1) - 12 = -6 < 0$$
• Since the second derivative is negative, $$f(x)$$ has a local maximum at $$x = 1$$.The local maximum value $$M$$ is:
  $$M = f(1) = (1)^3 - 6(1)^2 + 9(1) + 1 = 1 - 6 + 9 + 1 = 5$$
• At $$x = 3$$:
  $$f''(3) = 6(3) - 12 = 6 > 0$$
• Since the second derivative is positive, $$f(x)$$ has a local minimum at $$x = 3$$.The local minimum value $$m$$ is:
  $$m = f(3) = (3)^3 - 6(3)^2 + 9(3) + 1 = 27 - 54 + 27 + 1 = 1$$

Finally, compute the product $$Mm$$:

$$Mm = 5 \times 1 = 5$$