Finding The Range Of A Quadratic Function

Hey guys! Today, we're diving deep into the fascinating world of functions, specifically focusing on how to define the range of a function. We'll tackle a problem involving a quadratic function and explore its domain and range. Let's get started!

Understanding the Range of a Function

So, what exactly is the range of a function? In simple terms, the range is the set of all possible output values that a function can produce. Think of it as the "y-values" that your function spits out. When we talk about a function $f: x \rightarrow y$, the 'y' represents the codomain, and the range is the subset of the codomain that the function actually hits. For our problem, the function is defined as $f(x) = 2x^2 - 3x + 5$, and we're given a specific domain for $x$, which is the interval $[0, 2]$. This means we only care about the output values of the function when $x$ is between 0 and 2, inclusive.

Now, let's get to the core of the problem: finding the range of $f(x) = 2x^2 - 3x + 5$ where $x \in [0, 2]$. This function is a quadratic, meaning it's shaped like a parabola. The general form of a quadratic function is $ax^2 + bx + c$. In our case, $a = 2$, $b = -3$, and $c = 5$. Since the coefficient 'a' (which is 2) is positive, this parabola opens upwards. This is a crucial piece of information because it tells us that the function has a minimum value at its vertex. However, we need to consider the given domain, $[0, 2]$. The vertex might fall within this domain, or it might be outside it. If the vertex is within the domain, the minimum value of the function will occur at the vertex. If the vertex is outside the domain, the minimum and maximum values will occur at the endpoints of the domain.

To find the vertex of a parabola given by $f(x) = ax^2 + bx + c$, we use the formula for the x-coordinate of the vertex: $x_v = -b / (2a)$. In our function, $a = 2$ and $b = -3$. So, $x_v = -(-3) / (2 * 2) = 3 / 4$. Since $3/4$ is equal to $0.75$, and $0.75$ is within our domain $[0, 2]$, the minimum value of the function will indeed occur at the vertex.

To find the minimum value, we substitute $x = 3/4$ back into our function: $f(3/4) = 2(3/4)^2 - 3(3/4) + 5$. $f(3/4) = 2(9/16) - 9/4 + 5$ $f(3/4) = 18/16 - 9/4 + 5$ $f(3/4) = 9/8 - 18/8 + 40/8$ $f(3/4) = (9 - 18 + 40) / 8$ $f(3/4) = 31/8$

So, the minimum value of the function within the domain $[0, 2]$ is $31/8$.

Now, we need to find the maximum value. Since the parabola opens upwards and our domain is a closed interval, the maximum value must occur at one of the endpoints of the domain, $x = 0$ or $x = 2$. Let's evaluate the function at these endpoints:

For $x = 0$: $f(0) = 2(0)^2 - 3(0) + 5 = 0 - 0 + 5 = 5$.

For $x = 2$: $f(2) = 2(2)^2 - 3(2) + 5 = 2(4) - 6 + 5 = 8 - 6 + 5 = 7$.

Comparing the values at the endpoints ($5$ and $7$) and the value at the vertex ($31/8 = 3.875$), we see that the maximum value is $7$ (which occurs at $x=2$) and the minimum value is $31/8$ (which occurs at $x=3/4$).

Therefore, the range of the function $f(x) = 2x^2 - 3x + 5$ for $x \in [0, 2]$ is the interval $[31/8, 7]$. This means that all the possible output values of the function, when $x$ is between 0 and 2, lie between $31/8$ and $7$, inclusive. Understanding the range is super important in math, as it tells you the complete set of results you can expect from a given set of inputs for a specific function.

Proving Logical Equivalences with Truth Tables

Alright guys, moving on to a different, but equally cool, topic in logic: proving logical equivalences using truth tables. We're going to tackle the specific task of proving that the two conditional statements, $p \rightarrow q$ (read as "if p, then q") and $q \rightarrow p$ (read as "if q, then p"), are not logically equivalent. This second statement, $q \rightarrow p$, is actually called the converse of $p \rightarrow q$. We'll use a truth table to rigorously show this.

A truth table is a systematic way to list all possible combinations of truth values for the propositions involved and to determine the truth value of a compound statement for each combination. For our problem, we have two simple propositions, $p$ and $q$. Each proposition can be either true (T) or false (F). With two propositions, there are $2^2 = 4$ possible combinations of truth values:

1. $p$ is True, $q$ is True
2. $p$ is True, $q$ is False
3. $p$ is False, $q$ is True
4. $p$ is False, $q$ is False

Now, let's construct the truth table. We'll need columns for $p$, $q$, the conditional statement $p \rightarrow q$, and the conditional statement $q \rightarrow p$. Remember, a conditional statement $A \rightarrow B$ is only false when $A$ is true and $B$ is false. In all other cases, it's true.

Let's fill in the table:

Let's break down how we got these values:

• Row 1 (p=T, q=T):

  • $p \rightarrow q$: Since both $p$ and $q$ are true, the conditional is true. (T $\rightarrow$ T is T)
  • $q \rightarrow p$: Since both $q$ and $p$ are true, the conditional is true. (T $\rightarrow$ T is T)

• Row 2 (p=T, q=F):

  • $p \rightarrow q$: Here, $p$ is true and $q$ is false. This is the only case where a conditional statement is false. (T $\rightarrow$ F is F)
  • $q \rightarrow p$: Here, $q$ is false and $p$ is true. The conditional is true. (F $\rightarrow$ T is T)

• Row 3 (p=F, q=T):

  • $p \rightarrow q$: Here, $p$ is false and $q$ is true. The conditional is true. (F $\rightarrow$ T is T)
  • $q \rightarrow p$: Here, $q$ is true and $p$ is false. The conditional is false. (T $\rightarrow$ F is F)

• Row 4 (p=F, q=F):

  • $p \rightarrow q$: Since both $p$ and $q$ are false, the conditional is true. (F $\rightarrow$ F is T)
  • $q \rightarrow p$: Since both $q$ and $p$ are false, the conditional is true. (F $\rightarrow$ F is T)

Now, to determine if $p \rightarrow q$ and $q \rightarrow p$ are logically equivalent, we compare the truth values in the columns for $p \rightarrow q$ and $q \rightarrow p$. Two statements are logically equivalent if and only if they have the exact same truth value for every possible combination of truth values of their component propositions.

Looking at our truth table:

• In Row 1, both are T.
• In Row 2, $p \rightarrow q$ is F and $q \rightarrow p$ is T.
• In Row 3, $p \rightarrow q$ is T and $q \rightarrow p$ is F.
• In Row 4, both are T.

Since the truth values in the $p \rightarrow q$ column (T, F, T, T) are not identical to the truth values in the $q \rightarrow p$ column (T, T, F, T), the two conditional statements $p \rightarrow q$ and $q \rightarrow p$ are not logically equivalent. This demonstrates that the order in a conditional statement really matters! The implication going one way doesn't necessarily mean it holds true in the reverse direction. This is a fundamental concept in logic and is super useful when you're building arguments or analyzing statements. Keep practicing with these truth tables, guys; they're your best friend for proving logical relationships!

Conclusion

We've successfully navigated the process of finding the range of a quadratic function within a given domain and have rigorously proven, using a truth table, that a conditional statement and its converse are not logically equivalent. These are foundational concepts in mathematics that will serve you well as you continue your studies. Remember, practice makes perfect, so keep tackling those problems! Happy learning!