Problem 13
Question
Find all values of \(x\) such that \(f(x)>0\) and all \(x\) such that \(f(x)<0,\) and sketch the graph of \(f\) $$f(x)=\frac{1}{4} x^{3}-2$$
Step-by-Step Solution
Verified Answer
\( f(x) > 0 \) for \( x > 2 \), \( f(x) < 0 \) for \( x < 2 \).
1Step 1: Understanding the Function
We're given the function \( f(x) = \frac{1}{4}x^3 - 2 \). This is a cubic function shifted downward by 2 units. We need to determine when this function is greater than zero (\( f(x) > 0 \)) and when it is less than zero (\( f(x) < 0 \)).
2Step 2: Set the Function Equal to Zero
To find when \( f(x) = 0 \), we solve the equation \( \frac{1}{4}x^3 - 2 = 0 \). Adding 2 to both sides gives \( \frac{1}{4}x^3 = 2 \).
3Step 3: Solve for x
Multiply both sides by 4 to get rid of the fraction: \( x^3 = 8 \). Take the cube root of both sides to solve for \( x \): \( x = \sqrt[3]{8} = 2 \). Thus, \( f(x) = 0 \) when \( x = 2 \).
4Step 4: Determine when f(x) > 0
The function \( f(x) = \frac{1}{4}x^3 - 2 > 0 \) when \( \frac{1}{4}x^3 > 2 \). This simplifies to \( x^3 > 8 \), which implies \( x > 2 \).
5Step 5: Determine when f(x) < 0
The function \( f(x) = \frac{1}{4}x^3 - 2 < 0 \) when \( \frac{1}{4}x^3 < 2 \). This simplifies to \( x^3 < 8 \), which implies \( x < 2 \).
6Step 6: Sketch the Function
The function \( f(x) = \frac{1}{4}x^3 - 2 \) is a cubic curve shifted downwards by 2 units. It has an inflection point at the origin and crosses the x-axis at \( x = 2 \). The graph rises on the right (as \( x \rightarrow \infty \)) and falls on the left (as \( x \rightarrow -\infty \)). This indicates \( f(x) < 0 \) for \( x < 2 \) and \( f(x) > 0 \) for \( x > 2 \).
7Step 7: Confirm and Conclude
From our analysis, \( f(x) > 0 \) for \( x > 2 \) and \( f(x) < 0 \) for \( x < 2 \). The sketch confirms the behavior of the function.
Key Concepts
Graphing Cubic FunctionsUnderstanding Inequalities in Cubic FunctionsFunction Analysis of Cubic Functions
Graphing Cubic Functions
Graphing a cubic function involves understanding the basic shape and transformations that apply to the equation. For the function given, \[ f(x) = \frac{1}{4}x^3 - 2 \], the term \( \frac{1}{4}x^3 \) represents a standard cubic curve which is typically shaped like a stretched "S".
When you
Notice also the inflection point at the origin. This point indicates where the curvature of the graph changes direction. The function crosses the x-axis at the root \( x = 2 \). Study the parts where the function is above or below the x-axis to understand where it’s positive or negative.
When you
- Multiply by \( \frac{1}{4} \), it vertically compresses the function, making it less steep compared to \( x^3 \).
- Subtract 2, the entire graph shifts downward by 2 units, moving the curve below the x-axis initially.
Notice also the inflection point at the origin. This point indicates where the curvature of the graph changes direction. The function crosses the x-axis at the root \( x = 2 \). Study the parts where the function is above or below the x-axis to understand where it’s positive or negative.
Understanding Inequalities in Cubic Functions
Inequalities help us determine the intervals where a function is positive or negative. Given the cubic function \[ f(x) = \frac{1}{4}x^3 - 2 \], you will want to find where it is greater than zero, \( f(x) > 0 \), and less than zero, \( f(x) < 0 \).
To solve this:
To solve this:
- Set the function equal to zero and solve \( \frac{1}{4}x^3 - 2 = 0 \) to get \( x = 2 \). This tells us where the function crosses the x-axis.
- To find \( f(x) > 0 \), examine the inequality \( x^3 > 8 \), which simplifies to \( x > 2 \). This means the function is positive for any \( x \) greater than 2.
- For \( f(x) < 0 \), look at \( x^3 < 8 \), translating to \( x < 2 \). Therefore, the function is negative for any \( x \) less than 2.
Function Analysis of Cubic Functions
Analyzing a cubic function like \[ f(x) = \frac{1}{4}x^3 - 2 \] involves several key steps to understand its overall behavior and critical points.
First, consider transformations: the term \( -2 \) causes the overall function to shift down. This means the starting point of analysis considers this vertical shift.
Then, look for roots by solving \( f(x) = 0 \). The root \( x = 2 \) is crucial as it divides the function into regions of positivity and negativity.
First, consider transformations: the term \( -2 \) causes the overall function to shift down. This means the starting point of analysis considers this vertical shift.
Then, look for roots by solving \( f(x) = 0 \). The root \( x = 2 \) is crucial as it divides the function into regions of positivity and negativity.
- Before \( x = 2 \), the function decreases and remains negative. As you move from negative infinity towards \( x = 2 \), \( f(x) \) stays below the x-axis.
- After \( x = 2 \), \( f(x) \) becomes positive, indicating an increase as \( x \) heads towards positive infinity.
Other exercises in this chapter
Problem 13
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Sketch the graph of \(f\) $$f(x)=\frac{(4 x-1)(x-2)}{(2 x+3)(x-2)}$$
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Use the factor theorem to show that \(x-c\) is a factor of \(f(x)\). $$f(x)=x^{3}+x^{2}-2 x+12 ; \quad c=-3$$
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Show that the equation has no rational root. $$2 x^{5}+3 x^{3}+7=0$$
View solution