Question

The function $f(x) = x^3 - 6x^2 + 9x + 2$ has maximum value when $x$ is

• A. 1
• B. 2
• C. 3
• D.

Hint:

To find the local maximum or minimum of a function, first find its critical points by setting the first derivative to zero. Then, use the second derivative test: if $f''(x)<0$ at a critical point, it's a local maximum; if $f''(x)>0$, it's a local minimum.

Answer 1

The Correct Option is A

Step 1: Find the first derivative of the function $f(x)$.\n\[f(x) = x^3 - 6x^2 + 9x + 2\] \[f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 2) = 3x^2 - 12x + 9\]
Step 2: Set the first derivative to zero to find the critical points (potential locations of local maxima or minima).\n\[3x^2 - 12x + 9 = 0\] Divide by 3:\n\[x^2 - 4x + 3 = 0\] Factor the quadratic equation:\n\[(x-1)(x-3) = 0\] The critical points are $x=1$ and $x=3$.
Step 3: Find the second derivative of the function $f(x)$.\n\[f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12\]
Step 4: Evaluate the second derivative at each critical point to determine if it's a local maximum or minimum using the Second Derivative Test.\nAt $x=1$:\n\[f''(1) = 6(1) - 12 = 6 - 12 = -6\] Since $f''(1)<0$, the function has a local maximum at $x=1$.\n At $x=3$:\n\[f''(3) = 6(3) - 12 = 18 - 12 = 6\] Since $f''(3)>0$, the function has a local minimum at $x=3$.
Step 5: Conclude the point where the function has a maximum value.\nThe function $f(x)$ has a maximum value when $x=1$.