Intermediate Value Theorem: Real Zeros Of F(x)=5x^2-2x-6

Intermediate Value Theorem: Real Zeros of f(x)=5x^2-2x-6

Hey guys, let's dive into a super cool math problem today using the Intermediate Value Theorem (IVT). We're going to figure out if a specific function, $f(x)=5 x^2-2 x-6$, has at least one real zero between the values $a=-2$ and $b=-1$. This theorem is a real lifesaver when you want to know if a function crosses the x-axis within a given interval, even if you can't easily find the exact zero itself. So, buckle up, because we're about to put the IVT to the test!

Understanding the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus that basically says if a function is continuous on a closed interval $[a, b]$, and if $f(a)$ and $f(b)$ have opposite signs (meaning one is positive and the other is negative), then there must be at least one value $c$ between $a$ and $b$ where $f(c) = 0$. Think of it like this: if you're walking up a hill and then start walking down without ever jumping, you have to cross the exact same altitude level at some point on your path. The IVT works on the same principle for function values. For our problem, the function $f(x)=5 x^2-2 x-6$ is a polynomial. And guess what? All polynomials are continuous everywhere! This is a crucial piece of information because the IVT requires continuity. Since $f(x)$ is continuous on the interval $[-2, -1]$, we can proceed with applying the theorem. The main goal here is to evaluate the function at our endpoints, $a=-2$ and $b=-1$, and see if the results have different signs. If they do, boom, we know there's a zero in between. If they have the same sign, well, the IVT doesn't guarantee a zero, but it doesn't rule one out either. It just means the function might dip down and come back up, or vice versa, without crossing the x-axis in that specific interval. So, the strategy is straightforward: plug in the endpoints and check the signs of the outputs.

Evaluating the Function at the Endpoints

Alright, let's get down to business and evaluate our function $f(x)=5 x^2-2 x-6$ at the endpoints of our interval, $a=-2$ and $b=-1$. This is where the magic of the IVT starts to reveal itself. First up, we'll plug in $a=-2$ into our function:

$f(-2) = 5(-2)^2 - 2(-2) - 6$

Remember to follow the order of operations (PEMDAS/BODMAS, guys!).

$f(-2) = 5(4) - (-4) - 6$

$f(-2) = 20 + 4 - 6$

$f(-2) = 24 - 6$

$f(-2) = 18$

So, at $x=-2$, our function value is $18$. Now, let's move on to the other endpoint, $b=-1$. We'll substitute $x=-1$ into the function:

$f(-1) = 5(-1)^2 - 2(-1) - 6$

Again, let's be careful with the calculations.

$f(-1) = 5(1) - (-2) - 6$

$f(-1) = 5 + 2 - 6$

$f(-1) = 7 - 6$

$f(-1) = 1$

There we have it! At $x=-1$, our function value is $1$. Now that we have our results, $f(-2) = 18$ and $f(-1) = 1$, we can move to the next step of applying the Intermediate Value Theorem. This is where we look at the signs of these values. What do you notice about $18$ and $1$? They are both positive numbers, right? This is going to be key in determining whether we can conclude there's a real zero in the interval using the IVT.

Applying the Intermediate Value Theorem

Now comes the moment of truth, where we apply the Intermediate Value Theorem (IVT) to our findings. We've successfully calculated $f(-2) = 18$ and $f(-1) = 1$. The crucial part of the IVT states that if $f(a)$ and $f(b)$ have opposite signs, then there must be at least one real zero between $a$ and $b$. Let's examine the signs of our results. We found that $f(-2)$ is $18$, which is a positive number. We also found that $f(-1)$ is $1$, which is also a positive number. Since both $f(-2)$ and $f(-1)$ are positive, they do not have opposite signs. They are both on the same side of the x-axis (above it, in this case). This means that the condition required by the Intermediate Value Theorem to guarantee the existence of at least one real zero is not met. Therefore, based solely on the Intermediate Value Theorem, we cannot conclude that the function $f(x)=5 x^2-2 x-6$ has at least one real zero between $a=-2$ and $b=-1$. It's important to remember that the IVT only provides a guarantee when the signs are opposite. If the signs are the same, the theorem simply doesn't give us enough information to make a definitive statement about the existence of a zero within that interval. The function could cross the x-axis and come back, or it might just stay on one side of it. The IVT just doesn't tell us in this scenario. So, to answer the question directly: no, by the Intermediate Value Theorem, we cannot determine that there is at least one real zero between $a=-2$ and $b=-1$.

Conclusion: What Does This Mean?

So, guys, what's the final verdict? We used the Intermediate Value Theorem (IVT) to investigate the function $f(x)=5 x^2-2 x-6$ on the interval $[-2, -1]$. We found that $f(-2) = 18$ and $f(-1) = 1$. Since both of these values are positive, they do not have opposite signs. According to the Intermediate Value Theorem, if the function values at the endpoints of an interval have opposite signs, then there must be at least one real zero within that interval. Because our function values do not have opposite signs, the IVT does not guarantee that a real zero exists between $a=-2$ and $b=-1$. This means that, based on this theorem alone, we cannot determine that the function has at least one real zero in the specified interval. It's like checking if a path crosses a river by only looking at the height of the land at two points. If both points are higher than the river, you can't be sure if the path dips low enough to cross the river somewhere in between. However, this doesn't mean there isn't a zero there! It just means the IVT isn't the tool to prove its existence in this specific case. For a quadratic function like this, we could find the actual zeros using the quadratic formula, but the goal here was to practice the IVT. The takeaway is that the IVT is a powerful tool for proving the existence of zeros under specific conditions (continuous function, opposite signs at endpoints), but its inability to prove existence doesn't mean a zero isn't there. It just means we need a different approach or more information to confirm it. So, the answer to the question is: By the Intermediate Value Theorem, the function does not have at least one real zero (guaranteed). We can't say for sure using this method that there is one.

Let's Try Another Example (For Practice!)

To really solidify our understanding of the Intermediate Value Theorem (IVT), let's quickly look at a scenario where it would work. Imagine we have the function $g(x) = x^3 - x - 1$ and we want to know if it has a real zero between $a=1$ and $b=2$. First, $g(x)$ is a polynomial, so it's continuous everywhere, including the interval $[1, 2]$. Now, let's evaluate $g(x)$ at the endpoints:

$g(1) = (1)^3 - (1) - 1 = 1 - 1 - 1 = -1$

$g(2) = (2)^3 - (2) - 1 = 8 - 2 - 1 = 5$

Look at that! $g(1)$ is negative $(-1)$ and $g(2)$ is positive $(5)$. Since $g(1)$ and $g(2)$ have opposite signs, the Intermediate Value Theorem guarantees that there is at least one real zero for $g(x)$ between $x=1$ and $x=2$. This is the power of the IVT when its conditions are met! It gives us a definitive