Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Sketch the graph of \(f\). $$ f(x)=\frac{-3 x^{2}}{x^{2}+1} $$

Step-by-Step Solution

Answer

Graph is symmetric about the y-axis, has a max at (0,0), and approaches y=-3 as x goes to ±∞.

Step 1: Identify Domain

The function is given as \( f(x) = \frac{-3x^2}{x^2 + 1} \). The denominator \( x^2 + 1 \) is always positive and never zero for any real \( x \), so the domain of \( f \) is all real numbers, \( x \in \mathbb{R} \).

Step 2: Find Intercepts

For the y-intercept, set \( x = 0 \): \( f(0) = \frac{-3(0)^2}{0^2 + 1} = 0 \). So, the y-intercept is at \( (0, 0) \). Since the function is even, it is symmetric about the y-axis, and the x-intercept is also at \( (0,0) \).

Step 3: Determine Symmetry

Check for symmetry by testing \( f(-x) \). We have \( f(-x) = \frac{-3(-x)^2}{(-x)^2 + 1} = \frac{-3x^2}{x^2 + 1} = f(x) \). Thus, \( f(x) \) is even, indicating symmetry about the y-axis.

Step 4: Analyze Asymptotic Behavior

Find horizontal asymptotes by considering the limit as \( x \to \infty \) or \( x \to -\infty \). We have \( \lim_{x \to \pm\infty} f(x) = \lim_{x \to \pm \infty} \frac{-3x^2}{x^2 + 1} = -3 \). Thus, there is a horizontal asymptote at \( y = -3 \).

Step 5: Critical Points and Concavity

Find the derivative to locate critical points. \( f'(x) = \frac{d}{dx}\left(\frac{-3x^2}{x^2 + 1}\right) = \frac{-6x(x^2 + 1) - (-3x^2)(2x)}{(x^2 + 1)^2} = \frac{-6x}{(x^2 + 1)^2} \). Setting \( f'(x) = 0 \), we find \( x = 0 \) is a critical point. Use the second derivative, \( f''(x) = \frac{6(x^2 - 1)}{(x^2 + 1)^3} \), to test concavity. At \( x = 0 \), \( f''(0) = -6 \), indicating a local maximum at \( x = 0 \).

Step 6: Sketch the Graph

Using the information obtained: the graph has a local maximum point at \( (0,0) \), it is symmetric about the y-axis, and approaches \( y = -3 \) as \( x \to \pm \infty \). Draw a symmetric curve starting from the origin, bending downwards as it approaches the horizontal asymptote at \( y = -3 \) from above.

Key Concepts

Domain of a Function

The domain of a function refers to the set of all possible input values (usually represented by 'x') that allow the function to work without any issues. For the function given as \( f(x) = \frac{-3x^2}{x^2 + 1} \), the key factor in determining the domain is the denominator \( x^2 + 1 \). This expression is never zero because \( x^2 + 1 \) is always positive for any real number \( x \).
As a result, there are no restrictions or undefined points for this function, so the domain encompasses all real numbers. In mathematical terms, we express this as \( x \in \mathbb{R} \).
Understanding the domain is crucial because it tells us where the function exists and maintains continuous behavior.

Intercepts

Intercepts are the points where the graph of a function crosses the axes. For our function, we first look for the y-intercept. To find it, we set \( x = 0 \):

• Calculate the value of \( f(0) = \frac{-3(0)^2}{0^2 + 1} = 0 \). So, the y-intercept is at \( (0, 0) \).

Since the y-intercept calculation gives us the point where the graph crosses the y-axis, it also reveals that the graph passes through the origin. Moreover, because the function is even, meaning it is symmetric about the y-axis, the x-intercept coincides with the y-intercept, which is also at \( (0,0) \).
Intercepts are transition points for understanding how the function behaves near the origin.

Symmetry

Symmetry can make sketching a graph much simpler by reducing redundancy. A function is even if substituting \( -x \) for \( x \) gives the same function value, \( f(-x) = f(x) \). By testing symmetry for the given function:

• Calculate \( f(-x) = \frac{-3(-x)^2}{(-x)^2 + 1} = \frac{-3x^2}{x^2 + 1} = f(x) \).

This outcome confirms that the function is even. As a result, the graph of \( f \) is symmetric about the y-axis. Symmetry assists us in understanding how the function behaves in a mirrored fashion over the y-axis, simplifying the process of graph drawing.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as it approaches extreme values. Horizontal asymptotes concern what happens as \( x \) approaches infinity or negative infinity. For function \( f \), determining these limits involves:

• Calculate \( \lim_{x \to \pm \infty} \frac{-3x^2}{x^2 + 1} = -3 \).

Therefore, there is a horizontal asymptote at \( y = -3 \). This implies the graph will level out and approach this value, but never touch it, as \( x \) extends further in either direction.
Understanding horizontal asymptotes helps predict the long-term behavior of the function on a graph, giving us valuable insights into the function’s range limits.

Critical Points

Critical points on a graph occur where the derivative equals zero or is undefined, helping define maximum, minimum, or saddle points. For our function, derive \( f \) to find critical points:

• The derivative is \( f'(x) = \frac{-6x}{(x^2 + 1)^2} \).
• Setting \( f'(x) = 0 \) gives us \( x = 0 \).

The only critical point occurs at \( x = 0 \). Critical points are essential for identifying core transitions in the function's slope, indicating changes between increasing and decreasing behaviors.

Concavity

Concavity indicates whether the graph is curving upwards or downwards. By analyzing the second derivative, \( f''(x) \), we find the concavity of our function:

• The second derivative is \( f''(x) = \frac{6(x^2 - 1)}{(x^2 + 1)^3} \).
• At \( x = 0 \), \( f''(0) = -6 \).

Since \( f''(0) < 0 \), the graph is concave down at \( x = 0 \), indicating a local maximum there. Concavity assessment is key for understanding how and where the function bends, giving us insights into the nature of maxima and minima.