To find the critical points of the function $f(x) = \frac{x^2}{x^2 + 4}$, we need to follow these steps:

Step 1: Find the first derivative of $f(x)$.

We'll use the quotient rule for differentiation, which states that if $f(x) = \frac{g(x)}{h(x)}$, then:

$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$

Here, $g(x) = x^2$ and $h(x) = x^2 + 4$.

1. $g'(x) = 2x$
2. $h'(x) = 2x$

Using the quotient rule:

$f'(x) = \frac{(2x)(x^2 + 4) - (x^2)(2x)}{(x^2 + 4)^2}$

Simplifying the numerator:

$f'(x) = \frac{2x(x^2 + 4) - 2x(x^2)}{(x^2 + 4)^2}$

$f'(x) = \frac{2x^3 + 8x - 2x^3}{(x^2 + 4)^2}$

$f'(x) = \frac{8x}{(x^2 + 4)^2}$

Step 2: Find the critical points.

Critical points occur where the first derivative is zero or undefined.

1. Setting $f'(x) = 0$:

$\frac{8x}{(x^2 + 4)^2} = 0$

The only way this fraction equals zero is when the numerator is zero. So, set $8x = 0$:

$x = 0$

2. Check where $f'(x)$ is undefined:

The denominator $(x^2 + 4)^2$ is always positive for all real $x$ (since $x^2 + 4 > 0$ for all $x$). Therefore, the derivative is never undefined.

Step 3: Conclusion

The only critical point occurs at $x = 0$.

Would you like details on determining whether this is a maximum, minimum, or neither?

Follow-up Questions:

1. How do we classify critical points using the second derivative?
2. Can we use the first derivative test to check for local maxima or minima?
3. How do we find intervals of increasing and decreasing behavior for this function?
4. What are the asymptotic behaviors of the function as $x \to \infty$?
5. What is the domain and range of the function $f(x) = \frac{x^2}{x^2 + 4}$?

Tip:

Critical points are where the function's slope becomes zero or undefined. This is often where the function changes direction (maximums or minimums), but not always—so further tests are needed to classify them.