Evaluate F(x) = 4x^2 + 1: A Step-by-Step Guide

In this article, we'll explore how to evaluate the function $f(x) = 4x^2 + 1$ for different inputs. This is a fundamental concept in algebra and calculus, and mastering it will help you understand more complex mathematical ideas. We'll break down each part of the question step-by-step, ensuring you grasp the underlying principles. Let's dive in!

a) Evaluating $f(5)$

To find the value of $f(5)$, we need to substitute $x$ with $5$ in the function $f(x) = 4x^2 + 1$. This means we replace every instance of $x$ with the number $5$. Here’s how it looks:

$f(5) = 4(5)^2 + 1$

First, we need to calculate $5^2$, which is $5 * 5 = 25$. Now we substitute this value back into the equation:

$f(5) = 4(25) + 1$

Next, we multiply $4$ by $25$, which gives us $100$. So the equation becomes:

$f(5) = 100 + 1$

Finally, we add $1$ to $100$ to get the result:

$f(5) = 101$

Therefore, the value of the function $f(x)$ when $x = 5$ is $101$. This straightforward substitution and simplification process is crucial for understanding how functions behave with different inputs. Remember, the key is to carefully replace the variable with the given value and follow the order of operations (PEMDAS/BODMAS) to arrive at the correct answer. Evaluating functions at specific points like this is a foundational skill that will be used repeatedly in various mathematical contexts. It's also a great way to check your understanding of the function's behavior and its graphical representation. Keep practicing with different functions and values to build your confidence and proficiency.

b) Evaluating $f(-5)$

Now, let's evaluate the function $f(x) = 4x^2 + 1$ when $x = -5$. This is similar to the previous part, but it's important to pay attention to the negative sign. We substitute $x$ with $-5$:

$f(-5) = 4(-5)^2 + 1$

First, we need to calculate $(-5)^2$. Remember that squaring a negative number results in a positive number because a negative times a negative is positive. So, $(-5)^2 = (-5) * (-5) = 25$. Now we substitute this value back into the equation:

$f(-5) = 4(25) + 1$

Next, we multiply $4$ by $25$, which gives us $100$. So the equation becomes:

$f(-5) = 100 + 1$

Finally, we add $1$ to $100$ to get the result:

$f(-5) = 101$

Therefore, the value of the function $f(x)$ when $x = -5$ is $101$. Notice that $f(5)$ and $f(-5)$ both yield the same result. This is because the $x^2$ term makes the function even, meaning $f(x) = f(-x)$ for all $x$. Recognizing this property can save time and prevent errors when evaluating functions. It's also a good practice to check for symmetry in functions, as it provides valuable insights into their behavior. Continue to practice with different negative values to solidify your understanding of how functions handle negative inputs. This skill is particularly important in calculus and other advanced math topics where dealing with negative numbers is common.

c) Evaluating $f(s)$

In this part, we need to evaluate the function $f(x) = 4x^2 + 1$ when $x = s$. Here, $s$ is simply a variable, just like $x$. So we replace every instance of $x$ with $s$:

$f(s) = 4(s)^2 + 1$

This simplifies to:

$f(s) = 4s^2 + 1$

There's nothing more we can simplify here, as $s$ is just a variable. The expression $4s^2 + 1$ represents the value of the function $f(x)$ when the input is $s$. This exercise highlights the general nature of functions: they can accept any valid input, whether it's a number or another variable. Understanding this concept is essential for working with functions in abstract terms. It's also important to recognize that the variable $s$ could represent any value, and the expression $4s^2 + 1$ would still be valid. This kind of evaluation is commonly used in algebraic manipulations and proofs. Practice substituting different variables into functions to build your understanding of this fundamental concept. It’s a key skill that will be invaluable as you progress in your mathematical studies.

d) Evaluating $f(x+h)$

Now, let's evaluate the function $f(x) = 4x^2 + 1$ when $x$ is replaced with $(x+h)$. This might look a bit more complex, but we follow the same principle: substitute $x$ with $(x+h)$ in the function:

$f(x+h) = 4(x+h)^2 + 1$

First, we need to expand $(x+h)^2$. Remember that $(x+h)^2 = (x+h)(x+h)$. Using the distributive property (also known as FOIL), we get:

$(x+h)(x+h) = x^2 + xh + hx + h^2 = x^2 + 2xh + h^2$

Now we substitute this back into the equation:

$f(x+h) = 4(x^2 + 2xh + h^2) + 1$

Next, we distribute the $4$ across the terms inside the parentheses:

$f(x+h) = 4x^2 + 8xh + 4h^2 + 1$

There are no more like terms to combine, so this is the final expression for $f(x+h)$. This type of evaluation is particularly important in calculus, especially when dealing with derivatives and difference quotients. Understanding how to substitute and simplify expressions like this is crucial for grasping more advanced concepts. The expression $4x^2 + 8xh + 4h^2 + 1$ represents the value of the function $f(x)$ when the input is $x+h$. This skill is essential for understanding rates of change and other fundamental calculus concepts. Keep practicing with different functions and expressions to build your proficiency in this area.

In conclusion, we have successfully evaluated the function $f(x) = 4x^2 + 1$ for various inputs: $f(5) = 101$, $f(-5) = 101$, $f(s) = 4s^2 + 1$, and $f(x+h) = 4x^2 + 8xh + 4h^2 + 1$. These examples demonstrate the fundamental process of substituting values into functions and simplifying the resulting expressions. Mastering these skills is crucial for success in algebra, calculus, and beyond. Keep practicing and exploring different functions to build your mathematical intuition and problem-solving abilities.

For further learning on functions, visit Khan Academy's Functions and Equations.