Problem 15 Find the turning points on \(y=f... [FREE SOLUTION] (2024)

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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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Problem 15 Find the turning points on \(y=f... [FREE SOLUTION] (31)

Most popular questions from this chapter

Plot the graph of \(y=\sqrt{(} 1+x)\) for integral values of \(x\) between \(-1\)and 4 . Use your graph to estimate the value of \(\frac{\mathrm{d} y}{\mathrm{~d} x}\)when \(x=2\).Find the equations of the tangents to \(y=x^{3}+3 x\) which are parallel to theline \(y=15 x+2\)Find the maximum displacement of a particle from a point \(\mathrm{O}\), if itsdisplacement \(s\) metres from \(\mathrm{O}\) after time \(t\) seconds is given by $$ s=2+3 t-t^{2} $$\(f(x) \equiv x+\frac{1}{x}\) (a) \(f(x)\) is stationary when \(x=-1\). (b) \(\frac{\mathrm{d}}{\mathrm{d} x} f(x)=1-\frac{1}{x^{2}}\). (c) \(y=f(x)\) has no turning points.If a piece of string of fixed length is made to enclose a rectangle, show thatthe enclosed area is greatest when the rectangle is a square.
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Problem 15 Find the turning points on \(y=f... [FREE SOLUTION] (2024)

FAQs

How do you find turning points? ›

A polynomial of degree n can have up to (n−1) turning points. The number of turning points can be found by differentiating the function and setting the derivative equal to zero which will then give the x coordinates of any turning points. The number of solutions found corresponds to the number of turning points.

How do you find the coordinates of the turning point of F? ›

The turning point of the function of the form f(x)=ax2+q is determined by examining the range of the function.
  1. If a>0, the graph of f(x) is a “smile” and has a minimum turning point at (0;q).
  2. If a<0, the graph of f(x) is a “frown” and has a maximum turning point at (0;q).

What are my turning points? ›

A turning point happens when something changes direction, and it causes you to make a choice. It's the moment you decide that you need to make a shift and respond differently. It could be that what was happening before is no longer what you want to happen in the future.

What are turning points? ›

turning point. noun. : a point (as in an action or situation) where an important change occurs.

What is the formula for the turning point of Y? ›

The turning point form, y=a(x−h)2+k, provides the most information about the transformations that have occurred to a parabola. a causes a dilation by a factor of a from the x-axis. If a<0 (negative) the graph is reflected over the x-axis.

How do you find the Y point? ›

How to Find X and Y Coordinates? To find the x and y coordinates of a given point, See what is its perpendicular distance from the y-axis and it is your x-coordinate (x). See what is its perpendicular distance from the x-axis and it is your y-coordinate (y).

What is the formula to calculate the turning? ›

Tc=lm÷l=100÷200=0.5(min)

0.5×60=30(sec)The answer is 30 sec.

What is the vertex formula? ›

What is the vertex formula? The vertex formula is used to find the vertex of a parabola. The formula to find the vertex is (h, k) = (-b/2a, -D/4a), where D = b2-4ac.

What is the turning point formula for Grade 10? ›

To obtain the turning point or vertex (h, k) of the parabola, we can transform this equation to the vertex form of the parabola: y = a(x – h)2 + k.

How to find the turning point of an exponential function? ›

Change the base of the exponential, then differentiate as normal (using the quotient rule) and solve the resulting expression equal to zero, if possible. Then check the sign of the derivative either side of any such point. If the derivative changes sign, that is the (or a) turning point.

How do you find turning points at a level? ›

Stationary Points & Turning Points
  1. Step 1: Find the first derivative f'(x)
  2. Step 2: Solve f'(x) = 0 to find the x-coordinates of the stationary points.
  3. Step 3: Substitute those x-coordinates into f(x) to find the corresponding y-coordinates.

How do you test for turning points? ›

To find what type of turning point it is, find the second derivative (i.e. differentiate the function you get when you differentiate the original function), and then find what this equals at the location of the turning points. If it's positive, the turning point is a minimum.

How do you find and classify turning points? ›

A stationary point is called a turning point if the derivative changes sign (from positive to negative, or vice versa) at that point. There are two types of turning point: A local maximum, the largest value of the function in the local region. A local minimum, the smallest value of the function in the local region.

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