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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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