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Questions and Answers
What transformation does the constant 'k' perform on the graph of the function $f(x) = x^3 - x^2 - 5x + k$?
What transformation does the constant 'k' perform on the graph of the function $f(x) = x^3 - x^2 - 5x + k$?
- Reflects the graph over the x-axis.
- Shifts the graph up or down. (correct)
- Stretches the graph horizontally.
- Compresses the graph vertically.
Given that $f(x) = x^3 - x^2 - 5x + k$, what is the correct derivative, $f'(x)$?
Given that $f(x) = x^3 - x^2 - 5x + k$, what is the correct derivative, $f'(x)$?
- $x^2 - 2x - 5$
- $3x^2 - x - 5$
- $x^3 - x^2 - 5x$
- $3x^2 - 2x - 5$ (correct)
Critical points of a function are only found where the derivative equals zero.
Critical points of a function are only found where the derivative equals zero.
False (B)
The derivative of a function is given by $f'(x) = 3x^2 - 2x - 5$. Which of the following values are the solutions to $f'(x) = 0$?
The derivative of a function is given by $f'(x) = 3x^2 - 2x - 5$. Which of the following values are the solutions to $f'(x) = 0$?
If the derivative of a function, $f'(x)$, changes from negative to positive at a critical point, what does this indicate about the function at that point?
If the derivative of a function, $f'(x)$, changes from negative to positive at a critical point, what does this indicate about the function at that point?
What is the purpose of using a sign chart when analyzing the derivative of a function?
What is the purpose of using a sign chart when analyzing the derivative of a function?
If $f'(x) > 0$ for $x < a$, $f'(x) < 0$ for $a < x < b$, and $f'(x) > 0$ for $x > b$, then 'a' represents a local minimum.
If $f'(x) > 0$ for $x < a$, $f'(x) < 0$ for $a < x < b$, and $f'(x) > 0$ for $x > b$, then 'a' represents a local minimum.
Given $f(x) = x^3 - x^2 - 5x + k$, to solve for $k$ if there's a relative maximum at $x = a$ with a value of $M$, we substitute $x = a$ into $f(x)$ and set it equal to ______.
Given $f(x) = x^3 - x^2 - 5x + k$, to solve for $k$ if there's a relative maximum at $x = a$ with a value of $M$, we substitute $x = a$ into $f(x)$ and set it equal to ______.
Given the function $f(x) = x^3 - x^2 - 5x + k$ has a relative maximum at $x = -1$, which calculation correctly uses this information to solve for $k$ if the relative maximum value is 6?
Given the function $f(x) = x^3 - x^2 - 5x + k$ has a relative maximum at $x = -1$, which calculation correctly uses this information to solve for $k$ if the relative maximum value is 6?
After substituting $x = -1$ into $f(x) = x^3 - x^2 - 5x + k$ and setting the result equal to 6, what simplified equation needs to be solved for $k$?
After substituting $x = -1$ into $f(x) = x^3 - x^2 - 5x + k$ and setting the result equal to 6, what simplified equation needs to be solved for $k$?
For the function $f(x) = x^3 - x^2 - 5x + k$, given that it has a relative maximum of 6 at $x = -1$, what is the value of $k$?
For the function $f(x) = x^3 - x^2 - 5x + k$, given that it has a relative maximum of 6 at $x = -1$, what is the value of $k$?
The value of 'k' that causes a vertical shift in a cubic function will affect the x-coordinates of the relative maxima and minima.
The value of 'k' that causes a vertical shift in a cubic function will affect the x-coordinates of the relative maxima and minima.
To find critical points, one must find where the first derivative of a function is either equal to zero or is ______.
To find critical points, one must find where the first derivative of a function is either equal to zero or is ______.
Match the function analysis step with its primary purpose:
Match the function analysis step with its primary purpose:
Explain the geometric significance of finding the value of x where $f'(x) = 0$ for a function $f(x)$.
Explain the geometric significance of finding the value of x where $f'(x) = 0$ for a function $f(x)$.
Which of the following statements accurately describes the use of the first derivative test?
Which of the following statements accurately describes the use of the first derivative test?
If a cubic function has only one real root for its derivative, it cannot have both a relative maximum and a relative minimum.
If a cubic function has only one real root for its derivative, it cannot have both a relative maximum and a relative minimum.
A point on a function where the derivative changes sign is referred to as a relative extremum. If $f'(x)$ changes from positive to negative at $x=c$, then $f(c)$ is a relative ______.
A point on a function where the derivative changes sign is referred to as a relative extremum. If $f'(x)$ changes from positive to negative at $x=c$, then $f(c)$ is a relative ______.
What condition is necessary for the quadratic equation resulting from taking the derivative of a cubic function, $f'(x) = 0$, to ensure that the original cubic function, $f(x)$, has two distinct relative extrema?
What condition is necessary for the quadratic equation resulting from taking the derivative of a cubic function, $f'(x) = 0$, to ensure that the original cubic function, $f(x)$, has two distinct relative extrema?
Suppose a function has a relative maximum where $x = a$. Explain why simply knowing $f(a)$ isn't enough to fully define the original function $f(x) = x^3 - x^2 - 5x + k$ in this context.
Suppose a function has a relative maximum where $x = a$. Explain why simply knowing $f(a)$ isn't enough to fully define the original function $f(x) = x^3 - x^2 - 5x + k$ in this context.
Flashcards
What does 'k' do in a function?
What does 'k' do in a function?
A constant that shifts the graph of a function up or down on the y-axis.
How to solve for 'k'?
How to solve for 'k'?
To find 'k', determine the x-value of the relative maximum, plug it into the original function, set equal to the given y-value, and solve for 'k'.
What is the derivative of f(x) = x³ - x² - 5x + k?
What is the derivative of f(x) = x³ - x² - 5x + k?
f'(x) = 3x² - 2x - 5
What are critical points?
What are critical points?
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How do you find critical points?
How do you find critical points?
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Factored form of 3x² - 2x - 5 = 0?
Factored form of 3x² - 2x - 5 = 0?
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What are the solutions to (3x - 5)(x + 1) = 0?
What are the solutions to (3x - 5)(x + 1) = 0?
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What is a sign chart?
What is a sign chart?
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What intervals are created by the critical points x = -1 and x = 5/3?
What intervals are created by the critical points x = -1 and x = 5/3?
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How does the sign of f'(x) change at a relative maximum?
How does the sign of f'(x) change at a relative maximum?
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Where does the relative maximum occur?
Where does the relative maximum occur?
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What is the value of k?
What is the value of k?
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Study Notes
Solving for k in a Cubic Function with a Given Relative Maximum
- The problem is to find the value of k for the function f(x) = x³ - x² - 5x + k such that it has a relative maximum of 6.
- The constant k shifts the graph of the function up or down.
- The aim is to determine the value of k that shifts the graph so the relative maximum has a y-value of 6.
Finding the Derivative and Critical Points
- The derivative of f(x) is f'(x) = 3x² - 2x - 5.
- Critical points occur where the derivative is zero or undefined.
- In this case, the derivative is defined for all real numbers, so we only need to find where it equals zero.
- Setting the derivative equal to zero: 3x² - 2x - 5 = 0.
Factoring the Quadratic Equation
- The quadratic equation can be factored as (3x - 5)(x + 1) = 0.
- Solving for x gives two critical points: x = 5/3 and x = -1.
Using a Sign Chart to Determine Relative Maxima
- A sign chart is used to determine whether each critical point is a relative maximum or minimum.
- The number line is divided into intervals based on the critical points: x < -1, -1 < x < 5/3, and x > 5/3.
- By testing values in each interval, determine the sign of the derivative in that interval:
- For x < -1, f'(x) is positive (increasing).
- For -1 < x < 5/3, f'(x) is negative (decreasing).
- For x > 5/3, f'(x) is positive (increasing).
- The derivative changes from positive to negative at x = -1, indicating a relative maximum.
Solving for k Using the Relative Maximum
- The relative maximum occurs at x = -1, and we know the value of the function at this point is 6.
- Plug x = -1 into the original function: f(-1) = (-1)³ - (-1)² - 5(-1) + k = 6.
- Simplify the equation: -1 - 1 + 5 + k = 6.
- Combine terms: 3 + k = 6.
- Solve for k: k = 6 - 3 = 3.
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