Problem 11
Question
Explain why the following four statements ask for the same information: i) Find the roots of \(f(x)=x^{3}-3 x-1\) ii) Find the \(x\) -coordinates of the intersections of the curve \(y=x^{3}\) with the line \(y=3 x+1\) iii) Find the \(x\) -coordinates of the points where the curve \(y=x^{3}-3 x\) crosses the horizontal line \(y=1\) iv) Find the values of \(x\) where the derivative of \(g(x)=\) \((1 / 4) x^{4}-(3 / 2) x^{2}-x+5\) equals zero.
Step-by-Step Solution
Verified Answer
All statements solve \( x^3 - 3x - 1 = 0 \).
1Step 1: Understand the Concept of Roots
To find the roots of the equation \( f(x) = x^3 - 3x - 1 \), we set \( f(x) = 0 \) and solve for \( x \). This means we are looking for the values of \( x \) where the function intersects the x-axis, i.e., where \( f(x) = 0 \).
2Step 2: Set Up Equation for Intersection Points
Statement ii requires finding where the curve \( y = x^3 \) intersects the line \( y = 3x + 1 \). We set the equations equal to each other: \( x^3 = 3x + 1 \), which simplifies to \( x^3 - 3x - 1 = 0 \). This is the same equation as in step 1, confirming the same roots.
3Step 3: Evaluate Intersection with Horizontal Line
In statement iii, we determine where the curve \( y = x^3 - 3x \) is equal to the line \( y = 1 \). By setting \( x^3 - 3x = 1 \) and rearranging, we get \( x^3 - 3x - 1 = 0 \), matching the equation from previous steps.
4Step 4: Differentiate and Find Critical Points
Statement iv involves the derivative of \( g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5 \). First, find the derivative \( g'(x) = x^3 - 3x - 1 \). We then solve \( g'(x) = 0 \), which is equivalent to finding the roots of \( x^3 - 3x - 1 = 0 \).
5Step 5: Final Step: Conclude the Equivalence of Statements
Each statement leads to solving the same polynomial: \( x^3 - 3x - 1 = 0 \). Whether by finding roots, intersection points, or setting a derivative to zero, they all fundamentally ask for the same value of \( x \).
Key Concepts
Roots of a FunctionIntersection PointsCritical PointsDerivative Equations
Roots of a Function
The roots of a function are the values of \(x\) where the function equals zero. In essence, they are the points at which the graph of the function crosses the x-axis. For our example, we are looking for the roots of the equation \(f(x) = x^3 - 3x - 1\).
To find these, we set the equation to zero: \(f(x) = 0\). This means solving \(x^3 - 3x - 1 = 0\).
The concept of finding roots is crucial as it reveals valuable information about the function's behavior. These solutions tell us where the function has possible turning points or where it changes sign. By determining the x-values where the function is zero, you unlock significant insight into the graph's possible shape and intersections.
To find these, we set the equation to zero: \(f(x) = 0\). This means solving \(x^3 - 3x - 1 = 0\).
The concept of finding roots is crucial as it reveals valuable information about the function's behavior. These solutions tell us where the function has possible turning points or where it changes sign. By determining the x-values where the function is zero, you unlock significant insight into the graph's possible shape and intersections.
Intersection Points
Finding intersection points involves identifying where two curves meet. In the context of our exercise, we're asked to find where the curve \(y = x^3\) intersects with the line \(y = 3x + 1\).
To figure this out, we set the two equations equal: \(x^3 = 3x + 1\). This simplifies to the familiar equation \(x^3 - 3x - 1 = 0\).
Intersection points are critical in understanding how different equations relate to one another graphically. These points show exactly where one curve meets or crosses another on the coordinate plane, providing insight into any symmetry or similarity in the systems.
To figure this out, we set the two equations equal: \(x^3 = 3x + 1\). This simplifies to the familiar equation \(x^3 - 3x - 1 = 0\).
Intersection points are critical in understanding how different equations relate to one another graphically. These points show exactly where one curve meets or crosses another on the coordinate plane, providing insight into any symmetry or similarity in the systems.
Critical Points
Critical points in a function are where the derivative is zero or undefined, meaning the slope of the tangent line to the function is horizontal. This can indicate potential maxima, minima, or points of inflection.
For the function \(g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5\), we need to find its derivative \(g'(x)\). Calculating this gives us \(g'(x) = x^3 - 3x - 1\).
We look for when this derivative equals zero, solving \(x^3 - 3x - 1 = 0\), to find critical points. These critical points are where the function changes its slope - either peaking or troughing. Recognizing and understanding these points is essential for analyzing the function's graph and dynamics.
For the function \(g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5\), we need to find its derivative \(g'(x)\). Calculating this gives us \(g'(x) = x^3 - 3x - 1\).
We look for when this derivative equals zero, solving \(x^3 - 3x - 1 = 0\), to find critical points. These critical points are where the function changes its slope - either peaking or troughing. Recognizing and understanding these points is essential for analyzing the function's graph and dynamics.
Derivative Equations
Derivative equations express the rate at which a function's value changes with respect to changes in its variable. It provides a snapshot of the function's slope at any given point.
Evaluating the derivative of \(g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5\) gives us \(g'(x) = x^3 - 3x - 1\).
By setting \(g'(x) = 0\), we aim to find points where the function's slope is zero, revealing its critical points. This process of deriving and solving is fundamental to calculus, offering insights into the growth and elasticity of functions. Understanding derivative equations helps predict the behavior of a function, particularly around peaks and troughs.
Evaluating the derivative of \(g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5\) gives us \(g'(x) = x^3 - 3x - 1\).
By setting \(g'(x) = 0\), we aim to find points where the function's slope is zero, revealing its critical points. This process of deriving and solving is fundamental to calculus, offering insights into the growth and elasticity of functions. Understanding derivative equations helps predict the behavior of a function, particularly around peaks and troughs.
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