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Magazine Chapter 5: Problem 60
Suppose that \(f\) is a polynomial of degree 3 and that \(f^{\prime \prime}(x) \neq 0\) at any of the stationary point. Then (a) \(f\) has exactly one stationary point. (b) \(f\) must have no stationary point. (c) \(f\) must have exactly 2 stationary points. (d) \(f\) has either 0 or 2 stationary points.
Short Answer
Expert verified
(d) \(f\) has either 0 or 2 stationary points.
Step by step solution
01
Find the general form of a cubic polynomial
A degree 3 (cubic) polynomial function can be written in the form:
\[f(x) = ax^3 + bx^2 + cx + d\]
where a, b, c, and d are constants and a 鈮� 0.
02
Determine the first derivative
To find the stationary points, we need to find the first derivative of the function. Using the power rule, the first derivative of the cubic polynomial is:
\[f'(x) = 3ax^2 + 2bx + c\]
03
Set the first derivative equal to 0 to find stationary points
Stationary points occur when the first derivative is equal to 0:
\(3ax^2 + 2bx + c = 0\)
04
Determine the second derivative
The second derivative helps us determine the nature of stationary points (minimum, maximum or inflection). Let's find the second derivative using the first derivative:
\[f''(x) = 6ax + 2b\]
However, we are given that the second derivative is not equal to 0 at any stationary point. So, we can write:
\(6ax + 2b \neq 0\)
05
Analyze and identify the option with the correct number of stationary points
Since we have a cubic polynomial, we know that the first derivative is quadratic. A quadratic equation can have at most two real roots.
(a) A cubic function cannot have exactly one stationary point because if there is one real root, there must be another conjugate root.
(b) A cubic function can have no real stationary points because the first derivative's quadratic equation can have no real roots if the discriminant is negative.
(c) A cubic function can have exactly 2 stationary points as the first derivative's quadratic equation can have two real roots.
(d) 0 and 2 here holds as a correct answer. A cubic function can have either 0 or 2 stationary points, as derived from option (b) and (c) details.
Based on the above analysis, the correct answer is:
(d) \(f\) has either 0 or 2 stationary points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stationary Points
Stationary points are vital in understanding the behavior of a function. They are points where the function's slope is zero, meaning the graph is flat at these points. For any given differentiable function, stationary points occur where the first derivative of the function equals zero.
When dealing with a cubic polynomial, such as one of degree 3, stationary points can tell us where the curve changes direction or has a flat region. It's essential to note that stationary points can be classified into different types:
When dealing with a cubic polynomial, such as one of degree 3, stationary points can tell us where the curve changes direction or has a flat region. It's essential to note that stationary points can be classified into different types:
- Local Maximum: The function reaches a peak at this point.
- Local Minimum: The function reaches a trough.
- Inflection Point: The curve transitions from concave up to concave down or vice versa.
First Derivative
The first derivative of a function reveals its slope at any given point. For a cubic polynomial, this derivative is key to finding stationary points. The first derivative of a cubic polynomial \(f(x) = ax^3 + bx^2 + cx + d\) can be determined using the power rule. As shown:
\[f'(x) = 3ax^2 + 2bx + c\]
This is a quadratic expression. To find stationary points, set this equation equal to zero. Solving \(f'(x) = 0\) reveals the values of \(x\) that correspond to stationary points. A quadratic equation, like \(3ax^2 + 2bx + c = 0\), can have up to two real roots. This implies a cubic polynomial can have either 0, 1, or 2 stationary points, depending on the discriminant of this quadratic.
\[f'(x) = 3ax^2 + 2bx + c\]
This is a quadratic expression. To find stationary points, set this equation equal to zero. Solving \(f'(x) = 0\) reveals the values of \(x\) that correspond to stationary points. A quadratic equation, like \(3ax^2 + 2bx + c = 0\), can have up to two real roots. This implies a cubic polynomial can have either 0, 1, or 2 stationary points, depending on the discriminant of this quadratic.
Second Derivative
The second derivative of a function offers insights into the concavity of the graph, helping to determine the nature of each stationary point, whether it is a local maximum, minimum, or a point of inflection. For the cubic polynomial \(f(x) = ax^3 + bx^2 + cx + d\), the second derivative is:
\[f''(x) = 6ax + 2b\]
By substituting the \(x\)-values from your stationary points into \(f''(x)\), you can determine the nature of each point:
\[f''(x) = 6ax + 2b\]
By substituting the \(x\)-values from your stationary points into \(f''(x)\), you can determine the nature of each point:
- If \(f''(x) > 0\) at a point, it indicates a local minimum.
- If \(f''(x) < 0\) at a point, it indicates a local maximum.
- If \(f''(x) = 0\), the point could be an inflection point or need further analysis.
Roots of Equations
Roots of an equation are the values of \(x\) that satisfy the equation. In the context of the first derivative of a cubic polynomial, the roots of the equation \(f'(x) = 0\) signify the \(x\)-values of the stationary points.
A quadratic equation, such as the derivative \(3ax^2 + 2bx + c = 0\), can be solved using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The nature of the roots (real or complex) depends on the discriminant, \(b^2 - 4ac\):
A quadratic equation, such as the derivative \(3ax^2 + 2bx + c = 0\), can be solved using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
The nature of the roots (real or complex) depends on the discriminant, \(b^2 - 4ac\):
- Two distinct real roots: The discriminant is positive, leading to two stationary points.
- One real root (repeated): The discriminant is zero, indicating a possibility of symmetry.
- No real roots: The discriminant is negative, showing no stationary points.
