Finding the Minimum Value of the Polynomial Function

To find the minimum value of the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 6$, we use calculus. This involves finding the first derivative, identifying critical points, and using the second derivative test.

Step 1: Calculate the First Derivative

First, find the derivative of $f(x)$ with respect to $x$, denoted as $f'(x)$.

$ f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2 + 6) $

$ f'(x) = 12x^3 + 12x^2 - 24x $

Step 2: Find Critical Points

Set the first derivative $f'(x)$ equal to zero to find the critical points where the function might have a minimum or maximum value.

$ 12x^3 + 12x^2 - 24x = 0 $

Factor out the common term $12x$:

$ 12x(x^2 + x - 2) = 0 $

Factor the quadratic expression:

$ 12x(x+2)(x-1) = 0 $

The critical points are the values of $x$ that make the equation true:

• $12x = 0 \implies x = 0$
• $x+2 = 0 \implies x = -2$
• $x-1 = 0 \implies x = 1$

The critical points are $x = -2$, $x = 0$, and $x = 1$.

Step 3: Calculate the Second Derivative

Find the second derivative of $f(x)$, denoted as $f''(x)$, to determine whether each critical point corresponds to a local minimum, maximum, or inflection point.

$ f''(x) = \frac{d}{dx}(12x^3 + 12x^2 - 24x) $

$ f''(x) = 36x^2 + 24x - 24 $

Step 4: Apply the Second Derivative Test

Evaluate the second derivative at each critical point:

• At $x = -2$: $f''(-2) = 36(-2)^2 + 24(-2) - 24 = 36(4) - 48 - 24 = 144 - 72 = 72$. Since $f''(-2) > 0$, there is a local minimum at $x = -2$.
• At $x = 0$: $f''(0) = 36(0)^2 + 24(0) - 24 = -24$. Since $f''(0) < 0$, there is a local maximum at $x = 0$.
• At $x = 1$: $f''(1) = 36(1)^2 + 24(1) - 24 = 36 + 24 - 24 = 36$. Since $f''(1) > 0$, there is a local minimum at $x = 1$.

Step 5: Evaluate the Function at Minimum Points

Calculate the value of the function $f(x)$ at the points where local minima occur ($x = -2$ and $x = 1$).

• For $x = -2$: $ f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 6 $ $ f(-2) = 3(16) + 4(-8) - 12(4) + 6 $ $ f(-2) = 48 - 32 - 48 + 6 $ $ f(-2) = -26 $
• For $x = 1$: $ f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 6 $ $ f(1) = 3 + 4 - 12 + 6 $ $ f(1) = 1 $

Compare the values at the local minima: $f(-2) = -26$ and $f(1) = 1$. The absolute minimum value of the function is the smaller of these values.

Conclusion

The minimum value of the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 6$ is $-26$.