Evaluating The Function F(x) = X/(x^2+1) For Various Inputs

by Alex Johnson 60 views

Let's dive into evaluating the function $f(x) = \frac{x}{x^2 + 1}$ for several different inputs. This exercise will help us understand how functions work and how to substitute various values and expressions into them. We'll go through each part step by step, making sure everything is clear and easy to follow. This is a fundamental concept in mathematics, and mastering it will set you up for success in more advanced topics. So, let's put on our thinking caps and get started!

(a) Evaluating f(0)

To evaluate $f(0)$, we need to substitute $x$ with $0$ in the function. So, wherever we see $x$, we'll replace it with $0$. This is a straightforward start and helps us get a feel for the function's behavior at a specific point. It's like checking the function's value at its origin, giving us a baseline for further evaluations.

Here's how we do it:

f(0)=002+1=00+1=01=0f(0) = \frac{0}{0^2 + 1} = \frac{0}{0 + 1} = \frac{0}{1} = 0

So, $f(0) = 0$. This tells us that when the input is $0$, the output of the function is also $0$.

(b) Evaluating f(7)

Next, let's evaluate $f(7)$. This time, we'll substitute $x$ with $7$ in the function. This will give us a different value, showing how the function changes with different inputs. Evaluating at $x=7$ helps us see the function's behavior for a positive input, further expanding our understanding.

Let's plug in the numbers:

f(7)=772+1=749+1=750f(7) = \frac{7}{7^2 + 1} = \frac{7}{49 + 1} = \frac{7}{50}

Thus, $f(7) = \frac{7}{50}$. This means when we input $7$ into the function, the output is $\frac{7}{50}$.

(c) Evaluating f(-7)

Now, let's find $f(-7)$. We'll substitute $x$ with $-7$ in the function. This will help us see how the function behaves with negative inputs. Understanding the function's symmetry or asymmetry is crucial, and evaluating at $x=-7$ helps us with this.

Here's the calculation:

f(−7)=−7(−7)2+1=−749+1=−750f(-7) = \frac{-7}{(-7)^2 + 1} = \frac{-7}{49 + 1} = \frac{-7}{50}

So, $f(-7) = -\frac{7}{50}$. Notice that this is the negative of $f(7)$, which indicates the function might have some symmetry properties.

(d) Evaluating f(-x)

Now we're getting into more interesting territory! We need to find $f(-x)$, which means substituting $x$ with $-x$ in the function. This will help us understand how the function behaves when the input is negated. This is a key step in determining whether the function is even, odd, or neither.

Let's do the substitution:

f(−x)=−x(−x)2+1=−xx2+1f(-x) = \frac{-x}{(-x)^2 + 1} = \frac{-x}{x^2 + 1}

So, $f(-x) = -\frac{x}{x^2 + 1}$. This is equal to $-f(x)$, which means the function is an odd function. An odd function has the property that $f(-x) = -f(x)$.

(e) Evaluating -f(x)

Next, let's find $-f(x)$. This means we need to take the negative of the entire function $f(x)$. This is a direct way to see the impact of negating the output of the function.

Here's how we calculate it:

−f(x)=−xx2+1-f(x) = -\frac{x}{x^2 + 1}

Thus, $-f(x) = -\frac{x}{x^2 + 1}$. As we saw in part (d), this is equal to $f(-x)$, confirming that the function is odd.

(f) Evaluating f(x+4)

Now, let's evaluate $f(x+4)$. This means we substitute $x$ with $(x+4)$ in the function. This type of substitution helps us understand how the function shifts horizontally. It's a bit more complex, but stick with it, and you'll get the hang of it.

Let's make the substitution:

f(x+4)=x+4(x+4)2+1f(x+4) = \frac{x+4}{(x+4)^2 + 1}

Now we need to expand the denominator:

f(x+4)=x+4x2+8x+16+1=x+4x2+8x+17f(x+4) = \frac{x+4}{x^2 + 8x + 16 + 1} = \frac{x+4}{x^2 + 8x + 17}

So, $f(x+4) = \frac{x+4}{x^2 + 8x + 17}$. This gives us the function's expression when the input is shifted by 4 units.

(g) Evaluating f(4x)

Let's find $f(4x)$. This means we substitute $x$ with $4x$ in the function. This will show us how the function behaves when the input is scaled. Scaling inputs can significantly change the function's behavior, and this step helps us visualize that.

Here's the substitution:

f(4x)=4x(4x)2+1=4x16x2+1f(4x) = \frac{4x}{(4x)^2 + 1} = \frac{4x}{16x^2 + 1}

So, $f(4x) = \frac{4x}{16x^2 + 1}$. This is the function's expression when the input is multiplied by 4.

(h) Evaluating f(x+h)

Finally, let's evaluate $f(x+h)$. This is a crucial step in calculus, as it's related to the definition of the derivative. We're substituting $x$ with $(x+h)$ in the function. Understanding this substitution is essential for grasping calculus concepts.

Let's substitute and simplify:

f(x+h)=x+h(x+h)2+1f(x+h) = \frac{x+h}{(x+h)^2 + 1}

Now we expand the denominator:

f(x+h)=x+hx2+2xh+h2+1f(x+h) = \frac{x+h}{x^2 + 2xh + h^2 + 1}

So, $f(x+h) = \frac{x+h}{x^2 + 2xh + h^2 + 1}$. This expression is a key component in finding the derivative of the function using the limit definition.

Conclusion

In this exercise, we evaluated the function $f(x) = \frac{x}{x^2 + 1}$ for various inputs. We found $f(0)$, $f(7)$, $f(-7)$, $f(-x)$, $-f(x)$, $f(x+4)$, $f(4x)$, and $f(x+h)$. Each evaluation gave us a deeper understanding of the function's behavior. We also identified that the function is an odd function, meaning $f(-x) = -f(x)$. Understanding these concepts is crucial for further studies in mathematics, especially in calculus.

For further reading on functions and their properties, you might find the resources at Khan Academy helpful. They offer comprehensive explanations and practice exercises on this topic.