Finding Max And Min Of F(x) = X / (x^2 + 9) On [-4, 5]

When we talk about finding the absolute maximum and minimum of a function over a specific interval, we're essentially looking for the highest and lowest points that the function reaches within that defined range. Think of it like looking at a landscape from a specific viewpoint – you want to find the tallest peak and the deepest valley within that area. For the function $f(x) = \frac{x}{x^2+9}$ and the interval $[-4, 5]$, this involves a few key steps rooted in calculus. The Extreme Value Theorem tells us that if a function is continuous on a closed interval, it must attain both an absolute maximum and an absolute minimum on that interval. Our function $f(x)$ is a rational function, and since the denominator $x^2+9$ is never zero (because $x^2$ is always non-negative, so $x^2+9$ is always at least 9), it's continuous everywhere. This means we are guaranteed to find our absolute extrema on $[-4, 5]$. The process usually involves finding the critical points of the function within the interval and then comparing the function's values at these critical points with the function's values at the endpoints of the interval. Critical points are where the derivative of the function is either zero or undefined. So, our first major task is to calculate the derivative of $f(x)$, which we'll denote as $f'(x)$. We'll use the quotient rule for differentiation, which states that if $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$. In our case, $g(x) = x$ and $h(x) = x^2+9$. The derivatives are $g'(x) = 1$ and $h'(x) = 2x$. Applying the quotient rule, we get: $f'(x) = \frac{(1)(x^2+9) - (x)(2x)}{(x^2+9)^2} = \frac{x^2+9 - 2x^2}{(x^2+9)^2} = \frac{9-x^2}{(x^2+9)^2}$. Now that we have the derivative, we need to find the critical points. These are the values of $x$ where $f'(x) = 0$ or where $f'(x)$ is undefined. Since the denominator $(x^2+9)^2$ is always positive (as $x^2+9$ is always positive), $f'(x)$ is defined for all real numbers. Therefore, the only critical points will occur where the numerator is zero: $9-x^2 = 0$. Solving this equation, we get $x^2 = 9$, which means $x = 3$ or $x = -3$. Both of these critical points, $x=3$ and $x=-3$, lie within our interval $[-4, 5]$. So, these are the points we need to consider along with the endpoints of the interval.

Evaluating the Function at Critical Points and Endpoints

The next crucial step in finding the absolute maximum and minimum of $f(x)=\frac{x}{x^2+9}$ over $[-4,5]$ is to evaluate the function at the critical points we found ($x=-3$ and $x=3$) and at the endpoints of the interval ($x=-4$ and $x=5$). This systematic evaluation allows us to compare the function's values and pinpoint the absolute highest and lowest. Let's start with the critical points:

• At $x = -3$: $f(-3) = \frac{-3}{(-3)^2+9} = \frac{-3}{9+9} = \frac{-3}{18} = -\frac{1}{6}$.
• At $x = 3$: $f(3) = \frac{3}{(3)^2+9} = \frac{3}{9+9} = \frac{3}{18} = \frac{1}{6}$.

Now, let's evaluate the function at the endpoints of the interval $[-4, 5]$:

• At $x = -4$ (left endpoint): $f(-4) = \frac{-4}{(-4)^2+9} = \frac{-4}{16+9} = \frac{-4}{25}$.
• At $x = 5$ (right endpoint): $f(5) = \frac{5}{(5)^2+9} = \frac{5}{25+9} = \frac{5}{34}$.

We now have a list of candidate values for the absolute maximum and minimum:

• $f(-3) = -\frac{1}{6}$
• $f(3) = \frac{1}{6}$
• $f(-4) = -\frac{4}{25}$
• $f(5) = \frac{5}{34}$

To determine which of these is the absolute maximum and which is the absolute minimum, we need to compare these fractional values. It can be helpful to convert them to decimals or find a common denominator, but often a visual comparison or understanding the relative magnitudes is sufficient. Let's consider the positive values first: $\frac{1}{6}$ and $\frac{5}{34}$. To compare $\frac{1}{6}$ and $\frac{5}{34}$, we can cross-multiply: $1 \times 34 = 34$ and $6 \times 5 = 30$. Since $34 > 30$, we know that $\frac{1}{6} > \frac{5}{34}$. Therefore, $\frac{1}{6}$ is the larger of the two positive values.

Now let's consider the negative values: $-\frac{1}{6}$ and $-\frac{4}{25}$. To compare $-\frac{1}{6}$ and $-\frac{4}{25}$, it's easier to compare their positive counterparts, $\frac{1}{6}$ and $\frac{4}{25}$. Cross-multiplying: $1 \times 25 = 25$ and $6 \times 4 = 24$. Since $25 > 24$, we know that $\frac{1}{6} > \frac{4}{25}$. This means that when we introduce the negative sign, the inequality reverses: $-\frac{1}{6} < -\frac{4}{25}$. Therefore, $-\frac{1}{6}$ is the smaller (more negative) of the two negative values.

Comparing all four values: The largest value is $\frac{1}{6}$ and the smallest value is $-\frac{1}{6}$.

Conclusion: Absolute Maximum and Minimum Found

After carefully evaluating the function $f(x) = \frac{x}{x^2+9}$ at its critical points and the endpoints of the interval $[-4, 5]$, we can definitively state the absolute maximum and absolute minimum values. We found the following values:

• $f(-3) = -\frac{1}{6}$
• $f(3) = \frac{1}{6}$
• $f(-4) = -\frac{4}{25}$
• $f(5) = \frac{5}{34}$

By comparing these values, we determined that the largest value obtained is $\frac{1}{6}$, which occurs at $x=3$. Thus, the absolute maximum value of the function $f(x)$ on the interval $[-4, 5]$ is $\frac{1}{6}$.

Conversely, the smallest value obtained is $-\frac{1}{6}$, which occurs at $x=-3$. Therefore, the absolute minimum value of the function $f(x)$ on the interval $[-4, 5]$ is $-\frac{1}{6}$.

It's interesting to note that the absolute maximum and minimum occur at the critical points and not at the endpoints. This highlights the importance of checking all potential locations for extrema. The function $f(x)=\frac{x}{x^2+9}$ is an odd function, meaning $f(-x) = -f(x)$. This symmetry often leads to maximum and minimum values occurring at points that are symmetric with respect to the origin, as we see here with $x=3$ and $x=-3$. The interval $[-4, 5]$ is not perfectly symmetric around the origin, but because the critical points $x=-3$ and $x=3$ fall within this interval, and they yield the most extreme values, they become our absolute maximum and minimum points. The structure of the function, with $x$ in the numerator and $x^2+9$ in the denominator, causes the function to increase as $x$ increases from negative values towards $3$, reach a peak at $x=3$, and then decrease. Similarly, it decreases as $x$ becomes more negative, reaching a low point at $x=-3$, and then starts to increase.

For further reading on calculus concepts like finding extrema, you can check out resources from Khan Academy or Paul's Online Math Notes.