Quartic Function Minima: F(x) = 3x^4 + 4x^3 - 12x^2

Unveiling the Minimum of a Quartic Function: f(x) = 3x⁴ + 4x³ - 12x²

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of calculus to explore a specific quartic function: $f(x) = 3x^4 + 4x^3 - 12x^2$. We've got a snippet of its graph to give us a visual clue, and our mission, should we choose to accept it, is to find the exact coordinates of the point $M$ where this function hits its absolute minimum. We'll also be tackling a related question about the values of $b$ for which the equation $f(x) = b$ has a specific number of distinct real roots. Get ready to flex those mathematical muscles, because this is going to be a fun ride!

Part A: Pinpointing the Minimum Value

Alright, let's get down to business, shall we? To find the minimum value of the function $f(x) = 3x^4 + 4x^3 - 12x^2$, we need to employ the trusty tools of calculus. Specifically, we're looking for critical points, which are the candidates for local maxima and minima. These occur where the first derivative of the function is either zero or undefined. Since $f(x)$ is a polynomial, it's differentiable everywhere, so we only need to find where $f'(x) = 0$.

First up, let's find the derivative, $f'(x)$. Using the power rule for differentiation, which states that rac{d}{dx}(x^n) = nx^{n-1}, we differentiate each term of $f(x)$:

• The derivative of $3x^4$ is $4 imes 3x^{4-1} = 12x^3$.
• The derivative of $4x^3$ is $3 imes 4x^{3-1} = 12x^2$.
• The derivative of $-12x^2$ is $2 imes (-12)x^{2-1} = -24x$.

So, our first derivative is $f'(x) = 12x^3 + 12x^2 - 24x$.

Now, we set $f'(x)$ to zero to find our critical points:

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

To solve this cubic equation, we can start by factoring out the greatest common factor, which is $12x$:

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

This gives us one immediate solution: $12x = 0$, which means $x = 0$.

Now we need to solve the quadratic equation inside the parentheses: $x^2 + x - 2 = 0$. We can factor this quadratic. We're looking for two numbers that multiply to -2 and add up to 1. Those numbers are 2 and -1. So, we can factor it as:

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

This gives us two more solutions: $x + 2 = 0 ightarrow x = -2$ and $x - 1 = 0 ightarrow x = 1$.

Therefore, our critical points are $x = -2$, $x = 0$, and $x = 1$. These are the x-values where the function might have a local minimum or maximum. To determine which is which, and importantly, to find the absolute minimum, we can use the Second Derivative Test or analyze the sign changes of the first derivative. Given the shape of a quartic function with a positive leading coefficient (like ours, $3x^4$), we expect it to have a general 'W' or 'U' shape, implying it will have at least one absolute minimum.

Let's use the Second Derivative Test. First, we need to find the second derivative, $f''(x)$, by differentiating $f'(x) = 12x^3 + 12x^2 - 24x$:

• The derivative of $12x^3$ is $3 imes 12x^{3-1} = 36x^2$.
• The derivative of $12x^2$ is $2 imes 12x^{2-1} = 24x$.
• The derivative of $-24x$ is $1 imes (-24)x^{1-1} = -24$.

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

Now, we evaluate $f''(x)$ at each of our critical points:

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

We have found two local minima, at $x = -2$ and $x = 1$. To find the absolute minimum, we need to compare the function's values at these two points. We substitute these x-values back into the original function $f(x) = 3x^4 + 4x^3 - 12x^2$:

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

• For $x = 1$: $f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2$ $f(1) = 3(1) + 4(1) - 12(1)$ $f(1) = 3 + 4 - 12$ $f(1) = -5$

Comparing the values, $f(-2) = -32$ and $f(1) = -5$. The smaller value is -32. Therefore, the absolute minimum value of the function $f$ occurs at $x = -2$, and the minimum value is -32.

The coordinates of the point $M$, where the minimum value of the function $f$ occurs, are $(-2, -32)$. Pretty neat, right? We've successfully used calculus to pinpoint the lowest point on our graph!

Part B: Analyzing Roots and the Value of 'b'

Now for the second part of the puzzle, guys! We're asked to state the values of $b hetas R$ for which the equation $f(x) = b$ has a specific number of distinct real roots. This part is all about understanding how the horizontal line $y=b$ intersects the graph of $y=f(x)$. The number of intersection points directly corresponds to the number of distinct real roots of the equation $f(x) = b$.

We already know quite a bit about our function $f(x) = 3x^4 + 4x^3 - 12x^2$. We found its critical points: local minima at $x=-2$ (where $f(-2) = -32$) and $x=1$ (where $f(1) = -5$), and a local maximum at $x=0$ (where $f(0) = 3(0)^4 + 4(0)^3 - 12(0)^2 = 0$). We also know that as $x ightarrow e 0$, $f(x) ightarrow e e e$.

Let's visualize this. The graph comes down from positive infinity, hits a minimum at $(-2, -32)$, goes up to a local maximum at $(0, 0)$, comes down again to a local minimum at $(1, -5)$, and then goes back up towards positive infinity.

Now, consider a horizontal line $y = b$. The number of times this line crosses the graph of $y=f(x)$ tells us how many real solutions the equation $f(x) = b$ has.

• If $b < -32$: The horizontal line $y = b$ will be below the absolute minimum point $(-2, -32)$. It will not intersect the graph at all. So, there are zero distinct real roots.

• If $b = -32$: The horizontal line $y = b$ will touch the graph exactly at the absolute minimum point $(-2, -32)$. This point is a minimum, so it's a