Evaluating Functions Step-by-Step At X = -3

by Sharif Sakr 44 views

Hey guys! Today, we're diving into the world of functions and how to evaluate them at a specific point. We're going to take a close look at four different functions and see what happens when we plug in $x = -3$. So, grab your calculators and let's get started!

a) Evaluating $f(x) = x^3 - 64$ at $x = -3$

When it comes to evaluating polynomial functions, it's all about direct substitution. In this case, we have the function $f(x) = x^3 - 64$, and we want to find the value of $f(-3)$. This means we're going to replace every instance of $x$ in the function with $-3$.

So, let's break it down step by step:

  1. Substitute $x$ with $-3$: $f(-3) = (-3)^3 - 64$
  2. Calculate the exponent: $(-3)^3 = -3 \times -3 \times -3 = -27$
  3. Substitute back into the equation: $f(-3) = -27 - 64$
  4. Perform the subtraction: $f(-3) = -91$

Therefore, the value of the function $f(x) = x^3 - 64$ at $x = -3$ is $-91$. It's as simple as that! Just remember to follow the order of operations (PEMDAS/BODMAS) and you'll be golden. This particular function is a cubic function, and understanding how to evaluate it is crucial for graphing and analyzing its behavior.

Polynomial functions, like this one, are fundamental in mathematics and have wide-ranging applications in fields like physics, engineering, and economics. The ability to evaluate them efficiently is a key skill. We've seen how direct substitution works, but it's important to note that as functions become more complex, the evaluation process might involve additional steps such as simplifying expressions or dealing with different types of operations.

Consider, for instance, if we had a slightly more complex polynomial like $f(x) = 2x^3 + x^2 - 5x + 10$. Evaluating this at $x = -3$ would still involve the same principle of substitution, but we'd have to be careful with the order of operations. We'd first calculate the powers, then perform the multiplications, and finally add and subtract the terms. In this context, mastering the basics of polynomial evaluation is like building a strong foundation for tackling more advanced concepts in algebra and calculus. Remember, practice makes perfect, so try evaluating different polynomial functions at various points to solidify your understanding. This will not only help you in your math courses but also equip you with a valuable skill for problem-solving in many other areas of life. So, keep practicing, and you'll become a pro at function evaluation in no time!

b) Evaluating $g(x) = |x^3 - 3x^2 + 3x - 1|$ at $x = -3$

Now, let's tackle a function with an absolute value: $g(x) = |x^3 - 3x^2 + 3x - 1|$. The absolute value, denoted by the vertical bars, means that we're interested in the magnitude (or distance from zero) of the expression inside, regardless of its sign. So, if the expression inside evaluates to a negative number, we'll take its positive counterpart. If it's already positive or zero, we leave it as is.

Here's how we evaluate $g(x)$ at $x = -3$:

  1. Substitute $x$ with $-3$: $g(-3) = |(-3)^3 - 3(-3)^2 + 3(-3) - 1|$
  2. Calculate the powers: $(-3)^3 = -27$ and $(-3)^2 = 9$
  3. Substitute back into the equation: $g(-3) = |-27 - 3(9) + 3(-3) - 1|$
  4. Perform the multiplications: $g(-3) = |-27 - 27 - 9 - 1|$
  5. Perform the additions and subtractions inside the absolute value: $g(-3) = |-64|$
  6. Take the absolute value: $g(-3) = 64$

Therefore, the value of the function $g(x) = |x^3 - 3x^2 + 3x - 1|$ at $x = -3$ is $64$. The key here is to remember to perform all the operations inside the absolute value first and then apply the absolute value at the very end. Absolute value functions are important because they represent distances and magnitudes, which are always non-negative quantities. They appear in various contexts, such as error analysis, optimization problems, and signal processing.

Understanding how to evaluate absolute value functions is crucial not only in mathematics but also in real-world applications. The absolute value essentially strips away the sign of a number, providing a measure of magnitude or distance. In the context of our function $g(x)$, we first evaluate the expression inside the absolute value bars and then take the absolute value of the result. This ensures that the output is always non-negative.

Let's consider why this is important. Imagine you're calculating the difference between two temperatures. The absolute value of the difference tells you how far apart the temperatures are, regardless of which one is higher. Similarly, in engineering, you might use absolute values to measure the deviation of a manufactured part from its specifications. The absolute value gives you the magnitude of the error, without regard to whether the part is too large or too small. In mathematics, absolute value functions can create interesting graphs with sharp corners and discontinuities, which are important to understand in calculus. The function inside the absolute value, $x^3 - 3x^2 + 3x - 1$, is actually a perfect cube, specifically $(x-1)^3$. Recognizing this pattern can simplify the evaluation process. However, even if you don't recognize the pattern, the step-by-step approach we used will always lead you to the correct answer. So, whether you're dealing with temperature differences, manufacturing errors, or complex mathematical functions, the concept of absolute value is a powerful tool to have in your arsenal. Keep practicing with different functions and values, and you'll become more comfortable and confident in your ability to work with absolute values.

c) Evaluating $r(x) = \sqrt{3 - 2x}$ at $x = -3$

Next up, we have a function with a square root: $r(x) = \sqrt{3 - 2x}$. When dealing with square roots, we need to be mindful of the domain of the function. The expression inside the square root (the radicand) must be non-negative, otherwise, we'll end up with an imaginary number. So, before we even start plugging in $x = -3$, let's make sure that $3 - 2x \geq 0$.

Let's see what happens when we substitute $x = -3$:

  1. Substitute $x$ with $-3$: $r(-3) = \sqrt{3 - 2(-3)}$
  2. Perform the multiplication inside the square root: $r(-3) = \sqrt{3 + 6}$
  3. Perform the addition inside the square root: $r(-3) = \sqrt{9}$
  4. Calculate the square root: $r(-3) = 3$

Great! We got a real number, which means $x = -3$ is in the domain of the function. The value of the function $r(x) = \sqrt{3 - 2x}$ at $x = -3$ is $3$. It's worth noting that the domain of a function is a crucial concept in mathematics. It tells us the set of all possible input values (x-values) for which the function is defined. In the case of square root functions, the radicand must be non-negative, as we saw. In other types of functions, such as rational functions (fractions), we need to make sure that the denominator is not equal to zero.

Understanding the domain of a function is crucial, especially when dealing with square roots. In our function, $r(x) = \sqrt{3 - 2x}$, the expression under the square root (the radicand) must be non-negative. If it were negative, we'd end up with the square root of a negative number, which is not a real number. So, let's take a moment to delve deeper into why this matters and how we can determine the domain.

To find the domain, we set the radicand greater than or equal to zero: $3 - 2x \geq 0$. Now, let's solve this inequality for $x$. First, we subtract 3 from both sides: $-2x \geq -3$. Next, we divide both sides by -2. Remember, when we divide or multiply an inequality by a negative number, we need to flip the inequality sign: $x \leq \frac3}{2}$. This tells us that the domain of $r(x)$ is all real numbers less than or equal to $\frac{3}{2}$. Since $-3$ is indeed less than $\frac{3}{2}$, it's in the domain, and our calculation of $r(-3) = 3$ is valid. But what if we tried to evaluate $r(x)$ at, say, $x = 2$? Let's see $r(2) = \sqrt{3 - 2(2) = \sqrt{3 - 4} = \sqrt{-1}$. This is not a real number, which confirms that 2 is not in the domain of $r(x)$. This highlights the importance of checking the domain before evaluating a function, especially when square roots or other restrictions are involved. Failing to do so can lead to incorrect or undefined results. So, always be mindful of the domain, and you'll be on the right track!

d) Evaluating $q(x) = \frac{3x + 1}{x^2 + 7x + 10}$ at $x = -3$

Lastly, let's evaluate a rational function: $q(x) = \frac{3x + 1}{x^2 + 7x + 10}$. Rational functions are fractions where the numerator and denominator are both polynomials. A key thing to remember with rational functions is that the denominator cannot be zero. If the denominator is zero, the function is undefined at that point. So, before we plug in $x = -3$, let's check if it makes the denominator zero.

  1. Substitute $x$ with $-3$ into the denominator: $(-3)^2 + 7(-3) + 10 = 9 - 21 + 10 = -2$

Okay, the denominator is not zero at $x = -3$, so we're good to proceed.

Now, let's evaluate the function:

  1. Substitute $x$ with $-3$: $q(-3) = \frac{3(-3) + 1}{(-3)^2 + 7(-3) + 10}$
  2. Simplify the numerator: $3(-3) + 1 = -9 + 1 = -8$
  3. Simplify the denominator (we already did this above): $(-3)^2 + 7(-3) + 10 = -2$
  4. Substitute the simplified values back into the fraction: $q(-3) = \frac{-8}{-2}$
  5. Simplify the fraction: $q(-3) = 4$

Therefore, the value of the function $q(x) = \frac{3x + 1}{x^2 + 7x + 10}$ at $x = -3$ is $4$. With rational functions, always remember to check for values that make the denominator zero, as these values are not in the domain of the function. These points are often referred to as vertical asymptotes, which are vertical lines that the graph of the function approaches but never crosses.

Rational functions require careful attention to the denominator, and as we've seen, checking for values that make it zero is essential. The denominator of our function is a quadratic expression, $x^2 + 7x + 10$. To find the values of $x$ that make the denominator zero, we need to solve the equation $x^2 + 7x + 10 = 0$. This can be done by factoring, using the quadratic formula, or completing the square. In this case, the quadratic expression factors nicely: $(x + 2)(x + 5) = 0$. This gives us two solutions: $x = -2$ and $x = -5$. These are the values that make the denominator zero, and therefore, they are not in the domain of the function $q(x)$. They correspond to vertical asymptotes on the graph of the function. Vertical asymptotes are vertical lines that the graph of the function approaches but never crosses. They occur at values of $x$ where the denominator is zero and the numerator is non-zero. Understanding vertical asymptotes is crucial for sketching the graph of a rational function and analyzing its behavior. In addition to vertical asymptotes, rational functions can also have horizontal or slant asymptotes, which describe the behavior of the function as $x$ approaches positive or negative infinity. So, when working with rational functions, remember to always check the denominator, find the vertical asymptotes, and consider the behavior of the function at extreme values of $x$. This will give you a comprehensive understanding of the function and its properties.

Conclusion

So, there you have it! We've successfully evaluated four different functions at $x = -3$. Remember, the key is to carefully substitute the value of $x$ into the function and follow the order of operations. Also, be mindful of the domain of the function, especially when dealing with square roots and rational functions. Keep practicing, and you'll become a function evaluation master in no time! This exercise underscores the fundamental process of function evaluation in mathematics. By systematically substituting a given value into a function, we can determine the corresponding output. This skill is essential for various applications, including graphing functions, solving equations, and modeling real-world phenomena.