Question

The minimum value of $2x^3 - 9x^2 + 12x + 4$ is:

• A. $4$
• B. $5$
• C. $6$
• D. $7$
• E. $8$

Hint:

For cubic functions, if there are two critical points, one will always be a local maximum and the other a local minimum. The local minimum occurs at the larger $x$ value when the leading coefficient is positive.

Answer 1

Answer: (E)

Concept: To find the minimum value of a function $f(x)$, we find its critical points by setting the first derivative $f'(x) = 0$ and then use the second derivative test $f''(x) > 0$ to confirm the minimum.

Step 1: Find the first derivative and critical points.
Let $f(x) = 2x^3 - 9x^2 + 12x + 4$. \[ f'(x) = 6x^2 - 18x + 12 \] Set $f'(x) = 0$: \[ 6(x^2 - 3x + 2) = 0 \quad \Rightarrow \quad 6(x-1)(x-2) = 0 \] The critical points are $x = 1$ and $x = 2$.

Step 2: Apply the second derivative test.
\[ f''(x) = 12x - 18 \] At $x = 1$: $f''(1) = 12(1) - 18 = -6 < 0$ (Local Maximum). At $x = 2$: $f''(2) = 12(2) - 18 = 6 > 0$ (Local Minimum).

Step 3: Calculate the minimum value.
Substitute $x = 2$ back into $f(x)$: \[ f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 4 \] \[ f(2) = 2(8) - 9(4) + 24 + 4 = 16 - 36 + 28 = 8 \]