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Chapter 5: Problem 15
Find the turning points on \(y=f(x)\). (a) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=3 x-2\). (b) \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) is a quadratic function of \(x\). (c) The equation \(\frac{\mathrm{d} y}{\mathrm{~d} x}=0\) has two real roots,\(x=0, \quad x=1\).
Short Answer
Expert verified
The turning points of the equation occur at x=0 x=1 direct verifying curvature tx
Step by step solution
01
- Understand the Relationship Between Derivatives
Given: (a) \(\frac{\text{d}^2 y}{\text{d} x^2} = 3x - 2 \), (b) \(\frac{\text{d} y}{\text{d} x}\) is a quadratic function, (c) \(\frac{\text{d} y}{\text{d} x} = 0\) has roots \(x = 0\) and \(x = 1\).
02
- Find the First Derivative
Integrate the second derivative \(\frac{\text{d}^2 y}{\text{d} x^2} = 3x - 2\) to find the first derivative \(\frac{\text{d} y}{\text{d} x}\).\[\frac{\text{d} y}{\text{d} x} = \int (3x - 2) \text{d}x = \frac{3}{2}x^2 - 2x + C\]
03
- Use Information from the Problem
Since the first derivative is quadratic and it has roots at \(x = 0\) and \(x = 1\), it follows that:\[\frac{\text{d} y}{\text{d} x} = k(x)(x-1)\]where \(k\) is a constant.
04
- Determine the Constant
Equate the two expressions for \(\frac{\text{d} y}{\text{d} x}\). From Step 2: \(\frac{\text{d} y}{\text{d} x} = \frac{3}{2}x^2 - 2x + C \) and in Step 3: \(\frac{\text{d} y}{\text{d} x} = kx(x-1)\).We need to match these forms: \(kx^2 - kx = \frac{3}{2}x^2 - 2x + C\), which results in:
05
- Find Turning Points
The turning points occur where \(\frac{\text{d} y}{\text{d} x} = 0\).Set \(\frac{3}{2}x(x-1) = 0\). Solving for x shows points:
06
Finalizing
Turning points since second-fundamental further characterized solving original
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
second derivative test
The second derivative test helps us determine the nature of turning points (local maxima, minima, or points of inflection) on a function using the second derivative, denoted as \( \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} \).
When \( \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} > 0 \) at a critical point, the function has a local minimum there.
Alternatively, if \( \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} < 0 \), the function exhibits a local maximum.
A zero value might mean a point of inflection, but further analysis is needed.
In our problem:
Given \( \frac{\text{d}^2 y}{\text{d} x^2} = 3x - 2 \)
We need to check the values at the turning points \( x = 0 \) and \( x = 1 \).
- For \( x = 0 \): \( \frac{\text{d}^2 y}{\text{d} x^2} = 3(0) - 2 = -2 \), indicating a local maximum.
- For \( x = 1 \): \( \frac{\text{d}^2 y}{\text{d} x^2} = 3(1) - 2 = 1 \), indicating a local minimum.
integration
Integration is the process of finding the antiderivative or the area under the curve.
To find the first derivative \( \frac{\text{d} y}{\text{d} x} \) from \( \frac{\text{d}^2 y}{\text{d} x^2} \), we integrate:
Given \( \frac{\text{d}^2 y}{\text{d} x^2} = 3x - 2 \), we integrate:
\[ \frac{\text{d} y}{\text{d} x} = \int (3x - 2)\,dx = \frac{3}{2}x^2 - 2x + C \] where \( C \) is a constant that we determine using initial conditions or additional information.
quadratic functions
Quadratic functions are polynomial functions of degree 2, typically in the form \( ax^2 + bx + c \).
In this exercise, \( \frac{\text{d} y}{\text{d} x} \) is a quadratic function.
We derived \( \frac{\text{d} y}{\text{d} x} \) as \( \frac{3}{2}x^2 - 2x + C \).
Since we know the roots (zero points) of this function are \( x = 0 \) and \( x = 1 \), we express it as:
\[ \frac{\text{d} y}{\text{d} x} = kx(x-1) \] To match this to our derived equation and find \( k \), we solve:
\( kx^2 - kx = \frac{3}{2}x^2 - 2x + C \).
roots of equations
Roots of equations are the values of \( x \) that make an equation equal to zero.
For \( \frac{\text{d} y}{\text{d} x} = 0 \), solving this quadratic equation gives the roots of the function.
Given \( \frac{\text{d} y}{\text{d} x} = \frac{3}{2}x^2 - 2x + C \) with roots \( x = 0 \) and \( x = 1 \), we identify crucial points to further analyze.
These roots indicate potential turning points when the first derivative is zero.
We analyze these roots using the second derivative test to determine if they are maxima, minima, or points of inflection.
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