Chapter 0: Problem 29
Sketch a graph of the function showing all extreme, intercepts and asymptotes. $$f(x)=\frac{3 x}{\sqrt{x^{2}+4}}$$
Short Answer
Expert verified
The function intercepts the origin (0,0), has no extreme points or vertical asymptote, and has horizontal asymptotes at lines y=3 and y=-3.
Step by step solution
01
Understand the Function
The given function is \( f(x) = \frac{3 x}{\sqrt{x^{2}+4}} \). This is a rational function, a ratio where both the numerator and denominator are polynomials, where the denominator is the square root of a quadratic equation.
02
Find the Intercepts
The x intercept is found by setting \( f(x) = 0 \). This implies that \( x = 0 \). The y intercept is found by setting \( x = 0 \). So, \( y = f(0) = 0 \). Therefore, the function intercepts the origin (0,0) .
03
Find the Extreme Points
The extreme points for this function are at the critical points or undefined points. Therefore, we find the derivative \( f'(x) = \frac{3[x^{2}+4-(2x^{2})]}{(x^{2}+4)^{2}} \). Solving \( f'(x) = 0 \) gives no real solutions - this function does not have any extreme points.
04
Find the Asymptotes
The vertical asymptote is the value of x where the function does not exist or goes to ± infinity. But, the function exists for all x values, therefore it does not have any vertical asymptotes.However, to find the horizontal asymptote, we need to take the limit of the function as x approaches to positive and negative infinity. Both of these limits give us y=3. \( \lim_{{x \to \infty}} f(x) = 3 \) and \( \lim_{{x \to -\infty}} f(x) = -3 \), therefore this function has a horizontal asymptote at y=3 and y=-3.
05
Sketch the Graph
Now, plot the intercept, the limit values, and draw curves connecting these values and the limit lines at y=3 and y=-3. This will help sketch the graph of the function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Extreme Points
When discussing extreme points in a graph of a rational function, we're talking about local minima or maxima. These are points where the function reaches a lowest or highest value in its immediate vicinity. For our function, we found that solving the derivative \( f'(x) = 0 \) resulted in no real solutions. This means there are no extreme points in the graph.
The derivative simplifies as follows:
The derivative simplifies as follows:
- Calculate \( f'(x) = \frac{3(x^{2} + 4) - 6x^{2}}{(x^{2} + 4)^{2}} \).
- This reduces to \( \frac{3 \times 4 - 3x^{2}}{(x^{2} + 4)^{2}} \).
- Setting \( 12 - 3x^{2} = 0 \) yields no solution for real x.
Intercepts
Finding intercepts is a fundamental step when graphing rational functions. These intercepts tell us where the graph crosses the x-axis and y-axis.
For the function \( f(x) = \frac{3x}{\sqrt{x^{2}+4}} \):
For the function \( f(x) = \frac{3x}{\sqrt{x^{2}+4}} \):
- X-intercept: To find the x-intercept, set \( f(x) = 0 \). Here, the only solution is \( x = 0 \), as zero in the numerator makes the whole function zero.
- Y-intercept: For the y-intercept, set \( x = 0 \). Substituting this into the function gives \( f(0) = 0 \). Thus, the only point of intersection here is the origin, (0,0).
Asymptotes
Asymptotes in the graph of a rational function represent invisible boundaries the function approaches but never quite reaches. In this exercise, there are no vertical asymptotes because the function exists for all real values of \( x \); there are no x-values that make the denominator zero.
However, horizontal asymptotes can be identified by examining the values the function approaches as \( x \) moves towards positive or negative infinity.
However, horizontal asymptotes can be identified by examining the values the function approaches as \( x \) moves towards positive or negative infinity.
- Horizontal Asymptote: The horizontal asymptotes are found by calculating the limits \( \lim_{{x \to \infty}} f(x) = 3 \) and \( \lim_{{x \to -\infty}} f(x) = -3 \). So, the function approaches two horizontal lines: \( y = 3 \) and \( y = -3 \).