Unlock The Secrets: Solutions For F(x)=2x^3-x^2+8x-4

Hey there, math enthusiasts! Today, we're diving deep into a super interesting cubic function: $f(x)=2 x^3-x^2+8 x-4$. We're going to figure out what's true about its solutions. This isn't just about crunching numbers; it's about understanding the behavior of functions and the nature of their roots. Let's get this party started!

Grasping the Nature of Cubic Functions

Alright, guys, let's talk about cubic functions in general. A cubic function is a polynomial function of degree three. This means the highest power of $x$ in the function is 3. Functions like $f(x)=2 x^3-x^2+8 x-4$ fall into this category. Now, what's cool about cubic functions is that they have some predictable behaviors. For starters, their graphs always extend infinitely in both upward and downward directions. This is because as $x$ gets really, really big (either positive or negative), the $x^3$ term dominates the function's value. So, whether it goes up to the right and down to the left, or up to the left and down to the right, it's always going to cover all possible y-values.

Another key characteristic is that cubic functions always have at least one real root. Remember, a root (or a solution) is where the function's graph crosses or touches the x-axis, meaning $f(x) = 0$. Because the graph goes from negative infinity to positive infinity (or vice versa) on the y-axis, it has to cross the x-axis at least once. It might cross it three times, or it might cross it once and touch it at another point (which counts as a repeated root), but a minimum of one real root is guaranteed. This is a fundamental property stemming from the Intermediate Value Theorem, which basically says that if a continuous function takes on two different values, it must take on every value in between. Since our cubic function goes from negative to positive y-values (or vice versa), it must hit zero somewhere.

So, when we look at $f(x)=2 x^3-x^2+8 x-4$, we already know from the get-go that it's going to have at least one real solution. This is a crucial piece of information that helps us eliminate certain possibilities right off the bat when analyzing the given statements. Thinking about the graphical representation, we can anticipate that the graph of this function will intersect the x-axis at least once. This intersection point represents a real value of $x$ for which $f(x)$ equals zero. Understanding this general behavior of cubic polynomials sets a solid foundation for dissecting the specific properties of our given function and evaluating the provided statements with confidence. It's like having a map before you start your journey – you know the general terrain you'll encounter, which makes navigating the details much easier and more strategic. This upfront knowledge saves us a lot of time and helps us focus on the more nuanced aspects of the problem.

Analyzing the Function $f(x)=2 x^3-x^2+8 x-4$

Now, let's get down to business with our specific function: $f(x)=2 x^3-x^2+8 x-4$. We need to determine which statement about its solutions is true. We've already established that there must be at least one real solution. This immediately tells us that statement A, which claims there are no real solutions, is incorrect. The graph of a cubic function must cross the x-axis at least once, so the premise of statement A is flawed. It's important to trust these fundamental properties of polynomial functions; they're not just random rules, but consequences of how these functions behave.

Let's think about the possibility of complex solutions. Polynomials can have complex roots, which come in conjugate pairs if the polynomial has real coefficients (which ours does). Our function is of degree 3, so it must have exactly three roots in total, counting multiplicity. These three roots can be: three real roots, or one real root and a pair of complex conjugate roots. We know for sure there's at least one real root. So, the question boils down to whether the other two roots are also real, or if they form a complex conjugate pair.

To find the roots, we often try to factor the polynomial or use numerical methods. Let's try factoring by grouping, as it's a common technique for cubic polynomials that might have simpler structures. We can group the terms like this: $(2 x^3-x^2) + (8 x-4)$. Now, let's factor out the greatest common factor from each group. From the first group, we can factor out $x^2$, giving us $x^2(2x-1)$. From the second group, we can factor out 4, giving us $4(2x-1)$. So, our function looks like this: $f(x) = x^2(2x-1) + 4(2x-1)$.

Notice that we have a common binomial factor, $(2x-1)$. We can factor this out: $f(x) = (2x-1)(x^2+4)$.

Now, to find the solutions (where $f(x)=0$), we set each factor equal to zero:

1. $2x-1 = 0$ $2x = 1$ $x = 1/2$

This is our first real solution! We found it through factoring, which is awesome.

2. $x^2+4 = 0$ $x^2 = -4$ $x = \pm\sqrt{-4}$ $x = \pm 2i$

Here we have our other two solutions! They are $x = 2i$ and $x = -2i$. These are complex conjugate roots.

So, the solutions to $f(x)=2 x^3-x^2+8 x-4$ are $x = 1/2$, $x = 2i$, and $x = -2i$. This gives us one real solution and a pair of complex conjugate solutions. This fits perfectly with the properties of cubic functions we discussed earlier: one real root guaranteed, and the remaining roots can be real or complex conjugate pairs. Our factoring revealed exactly this scenario.

Evaluating the Statements

With the solutions in hand ($x=1/2$, $x=2i$, $x=-2i$), let's revisit the statements and see which one holds true.

• Statement A: "There are no real solutions, because the function's graph does not cross the $x$-axis." We found a real solution, $x=1/2$. Also, cubic functions always cross the x-axis at least once. So, this statement is definitely false. Our analysis showed a clear real root.

• Statement B: "$x=-i$ is a solution, because..." Let's check if $x=-i$ is a solution. If we plug it into the function: $f(-i) = 2(-i)^3 - (-i)^2 + 8(-i) - 4$ Recall that $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, etc. $f(-i) = 2(-i) - (-1) + (-8i) - 4$ $f(-i) = -2i + 1 - 8i - 4$ $f(-i) = -3 - 10i$ Since $f(-i)$ is not equal to 0, $x=-i$ is not a solution. So, statement B, in its entirety (including whatever reason it might give), must be false because the premise that $x=-i$ is a solution is incorrect. We found the complex solutions to be $2i$ and $-2i$, not $-i$.

• Statement C: "$x=2i$ is a solution, because the factor $(x^2+4)$ results from factoring the polynomial." We found that $f(x) = (2x-1)(x^2+4)$. When we set $x^2+4=0$, we get $x^2 = -4$, which yields $x = \pm 2i$. Therefore, $x=2i$ is indeed a solution. The reason provided is also correct: the factor $(x^2+4)$ arises directly from factoring the polynomial, and setting this factor to zero gives us the complex roots. This statement appears to be true.

• Statement D: "$x=-1/2$ is a solution, because..." Let's test $x=-1/2$ in our original function: $f(-1/2) = 2(-1/2)^3 - (-1/2)^2 + 8(-1/2) - 4$ $f(-1/2) = 2(-1/8) - (1/4) + (-4) - 4$ $f(-1/2) = -1/4 - 1/4 - 4 - 4$ $f(-1/2) = -2/4 - 8$ $f(-1/2) = -1/2 - 8$ $f(-1/2) = -17/2$ Since $f(-1/2)$ is not 0, $x=-1/2$ is not a solution. So, statement D is false. Our real solution was $x=1/2$, not $x=-1/2$.

Conclusion: The True Statement

Based on our thorough analysis and factoring, the only true statement about the solutions of $f(x)=2 x^3-x^2+8 x-4$ is that $x=2i$ is a solution, because the factor $(x^2+4)$ results from factoring the polynomial. This statement correctly identifies one of the complex roots and provides a valid reason derived from the structure of the function itself. It's awesome how breaking down the polynomial into its factors ($ (2x-1)(x^2+4) $) directly leads us to all the roots: one real ($1/2$) and two complex conjugates ($2i$ and $-2i$). Understanding these algebraic manipulations and their connection to the function's roots is key to mastering these types of problems. Keep practicing, guys, and you'll be factoring like pros in no time!