Problem 2
Question
(a) Sketch the graph of \(f\). (b) Find the domain \(D\) and range \(R\) of \(f .\) (c) Find the intervals on which \(f\) is increasing or is decreasing. $$f(x)=\frac{1}{x^{2}}$$
Step-by-Step Solution
Verified Answer
The graph has two branches, domain is \( (-\infty, 0) \cup (0, \infty) \), range is \( (0, \infty) \), decreasing on \( (-\infty, 0) \) and \( (0, \infty) \).
1Step 1: Analyze Basic Graph Shape
The function is given by \( f(x) = \frac{1}{x^2} \). Before sketching, note that this is a reciprocal function flipped along the x-axis due to the negative exponent. It produces a curve with two branches, one in the first quadrant and one in the second quadrant.
2Step 2: Determine Asymptotes
The graph of \( f(x) = \frac{1}{x^2} \) has two asymptotes. A vertical asymptote at \( x = 0 \) since division by zero is undefined, meaning the function approaches infinity as \( x \to 0^+ \) and \( x \to 0^- \). The horizontal asymptote is \( y = 0 \), as \( x \to \pm \infty \), \( f(x) \to 0 \).
3Step 3: Sketch the Graph
Using the information from the previous steps, sketch the graph with branches in the first and second quadrants approaching the x-axis and diverging at \( x = 0 \).
4Step 4: Define the Domain of \( f \)
The domain of \( f(x) = \frac{1}{x^2} \) is all real numbers except \( x = 0 \) because the function is undefined at \( x = 0 \). Thus, the domain \( D \) is \( (-\infty, 0) \cup (0, \infty) \).
5Step 5: Define the Range of \( f \)
Since the function \( f(x) = \frac{1}{x^2} \) is always positive and decreases towards zero as \( |x| \) increases, the range \( R \) is \( (0, \infty) \).
6Step 6: Find Intervals of Increase and Decrease
For \( f(x) = \frac{1}{x^2} \), observe that as \( x \) moves from negative infinity to zero, \( f(x) \) decreases, and as \( x \) moves from just above zero to infinity, \( f(x) \) also decreases. Therefore, the function is decreasing on both intervals \( (-\infty, 0) \) and \( (0, \infty) \).
Key Concepts
Domain and RangeAsymptotes in FunctionsIncreasing and Decreasing Intervals
Domain and Range
The domain of a function represents all the possible values of the variable for which the function is defined. For the function \(f(x) = \frac{1}{x^2}\), the domain excludes \(x = 0\), where the function is undefined due to division by zero. Therefore, the domain is \((-\infty, 0) \cup (0, \infty)\).
The range of a function is the set of all possible output values. For \(f(x) = \frac{1}{x^2}\), the output is always positive since square of any real number is non-negative, making the range \((0, \infty)\).
Understanding domain and range is crucial in comprehending the limitations and behavior of any given function.
The range of a function is the set of all possible output values. For \(f(x) = \frac{1}{x^2}\), the output is always positive since square of any real number is non-negative, making the range \((0, \infty)\).
Understanding domain and range is crucial in comprehending the limitations and behavior of any given function.
- Domain: All possible \(x\)-values. For this function: \((-\infty, 0) \cup (0, \infty)\)
- Range: All possible \(f(x)\)-values. For this function: \((0, \infty)\)
Asymptotes in Functions
Asymptotes are lines that the graph of a function approaches but never actually meets. They provide significant insights into the behavior of the graph at extreme values. For \(f(x) = \frac{1}{x^2}\), there are two types of asymptotes to identify:
- **Vertical Asymptote:** This occurs at \(x = 0\) because the function becomes undefined as \(x\) approaches zero. As \(x\to 0^+\) or \(x\to 0^-\), \(f(x)\) tends towards infinity. Thus, there's a vertical asymptote along this line.
- **Horizontal Asymptote:** At extreme values of \(x\) (as \(x\to\pm \infty\)), the function \(f(x)\) tends towards zero, but never actually reaches zero, resulting in a horizontal asymptote at \(y = 0\).
Recognizing asymptotes informs how the graph stretches towards infinity and how the function behaves at its boundary values.
- **Vertical Asymptote:** This occurs at \(x = 0\) because the function becomes undefined as \(x\) approaches zero. As \(x\to 0^+\) or \(x\to 0^-\), \(f(x)\) tends towards infinity. Thus, there's a vertical asymptote along this line.
- **Horizontal Asymptote:** At extreme values of \(x\) (as \(x\to\pm \infty\)), the function \(f(x)\) tends towards zero, but never actually reaches zero, resulting in a horizontal asymptote at \(y = 0\).
Recognizing asymptotes informs how the graph stretches towards infinity and how the function behaves at its boundary values.
- Vertical Asymptote: \(x = 0\)
- Horizontal Asymptote: \(y = 0\)
Increasing and Decreasing Intervals
When analyzing a function, it's useful to determine where it's increasing or decreasing.
For \(f(x) = \frac{1}{x^2}\), as \(x\) moves from \(-\infty\) to 0, and from 0 to \(+\infty\), the function is decreasing. This means that as we move along the x-values, the output of the function gets smaller.
Remember:
- **Decreasing interval:** Both \((-\infty, 0)\) and \((0, \infty)\). The function decreases in both directions away from zero.
Understanding these intervals helps in predicting the trend of the function as \(x\) varies.
For \(f(x) = \frac{1}{x^2}\), as \(x\) moves from \(-\infty\) to 0, and from 0 to \(+\infty\), the function is decreasing. This means that as we move along the x-values, the output of the function gets smaller.
Remember:
- A function is increasing if the values of \(f(x)\) rise as \(x\) increases.
- A function is decreasing if the values of \(f(x)\) drop as \(x\) rises.
- **Decreasing interval:** Both \((-\infty, 0)\) and \((0, \infty)\). The function decreases in both directions away from zero.
Understanding these intervals helps in predicting the trend of the function as \(x\) varies.
Other exercises in this chapter
Problem 1
Find the quotient and remainder if \(f(x)\) is divided by \(p(x)\). $$f(x)=2 x^{4}-x^{3}-3 x^{2}+7 x-12 ; \quad p(x)=x^{2}-3$$
View solution Problem 2
Exer. 1-12: Express the statement as a formula that involves the given variables and a constant of proportionallty \(k,\) and then determine the value of \(k\)
View solution Problem 2
Find a polynomial \(f(x)\) of degree 3 that has the indicated zeros and satisfies the given condition. $$-5,2,4 ; \quad f(3)=-24$$
View solution Problem 2
Sketch the graph of \(f\) for the indicated value of \(c\) or \(a\) $$f(x)=-2 x^{3}+c$$ (a) \(c=-2\) (b) \(c=2\)
View solution