Problem 78
Question
Give reasons for your answers. Let \(f(x)=\left|x^{3}-9 x\right|\) a. Does \(f^{\prime}(0)\) exist? b. Does \(f^{\prime}(3)\) exist? c. Does \(f^{\prime}(-3)\) exist? d. Determine all extrema of \(f\)
Step-by-Step Solution
Verified Answer
a. Yes, \(f'(0)\) exists. b. No, \(f'(3)\) doesn't exist. c. No, \(f'(-3)\) doesn't exist. d. Extrema at \(x = 3\) and \(-3\) (local minima).
1Step 1: Analyze the function
Begin by discussing the core function, \(f(x) = |x^3 - 9x|\). The absolute value impacts differentiability, particularly around points where the expression inside goes to zero, i.e., \(x^3 - 9x = 0\). Solve \(x(x^2 - 9) = 0\) to find critical points. Solutions are \(x = 0, 3, -3\).
2Step 2: Check existence of \(f'(0)\)
To check if \(f'(0)\) exists, observe that \(f(x)\) can be written as \(f(x) = |x(x+3)(x-3)|\). Around \(x = 0\), \(f(x)\) simplifies to \(f(x) = x^3 - 9x\) and is differentiable. Compute the derivative: \(f'(x) = 3x^2 - 9\). Substitute \(x = 0\) to find \(f'(0) = -9\). Since a clear derivative exists, \(f'(0)\) exists and equals \(-9\).
3Step 3: Check existence of \(f'(3)\)
Examine whether \(f(x) = |x(x+3)(x-3)|\) is differentiable at \(x = 3\). Notice that at \(x = 3\), \(x^3 - 9x = 0\), leading to a potential corner or cusp in the graph after applying the absolute value, suggesting \(f(x) = 0\). Calculate right- and left-hand derivatives. Right-hand: \(f'(x) = 3x^2 - 9\), so \(f'(x) \to 18\) as \(x \to 3^+\). Left-hand: \(f'(x) \to -18\) as \(x \to 3^-\). Different one-sided derivatives imply \(f'(3)\) doesn't exist.
4Step 4: Check existence of \(f'(-3)\)
Similar to \(x = 3\), for \(x = -3\), the expression inside the absolute value equals zero, creating a potential non-differentiable point. Calculate right- and left-hand derivatives: Right-hand: \(f'(x) = 3x^2 - 9\) gives \(f'(x) \to 18\) as \(x \to -3^+\). Left-hand: \(f'(x) \to -18\) as \(x \to -3^-\). Since the side derivatives do not match, \(f'(-3)\) does not exist.
5Step 5: Determine extrema
Find potential extrema by analyzing critical points determined where the derivative may be zero: \(x = 0, 3, -3\). Compute \(f(x)\) at these points: \(f(0) = 0\), \(f(3) = 0\), \(f(-3) = 0\). Examine intervals \((-\infty, -3), (-3, 0), (0, 3), (3, \infty)\) using derivative test signs: Negative on \(-3\) to \(0\), positive on both sides of \(3\) and \(-3\). Thus, \(x = 3, -3\) are local minima, and there is no apparent local maximum.
Key Concepts
ExtremaAbsolute ValueCritical PointsDerivative Tests
Extrema
The concept of extrema in calculus refers to the points on a function where it takes a maximum or minimum value. Extrema are categorized into local and global types. A local extremum is a point where the function value is higher or lower than all nearby points. On the other hand, a global extremum is a point where the function value is the highest or lowest over the entire domain.
In the case of a function defined by the absolute value, such as \( f(x) = |x^3 - 9x| \), the extrema often occur at critical points where the derivative of the function is zero or undefined. In this problem, the critical points we found were \(x = 0, 3, -3\).
To determine whether these critical points are extrema, we check the sign of the derivative around these points. With derivative test signs, we found that \(x = 3\) and \(x = -3\) are local minima since these points make the derivative change from negative to positive.
In the case of a function defined by the absolute value, such as \( f(x) = |x^3 - 9x| \), the extrema often occur at critical points where the derivative of the function is zero or undefined. In this problem, the critical points we found were \(x = 0, 3, -3\).
To determine whether these critical points are extrema, we check the sign of the derivative around these points. With derivative test signs, we found that \(x = 3\) and \(x = -3\) are local minima since these points make the derivative change from negative to positive.
Absolute Value
Absolute value functions, such as \( f(x) = |x^3 - 9x| \), have specific features that impact differentiability and the appearance of the graph. Absolute values transform the graph of a function by reflecting any negative parts into positive values. This can create sharp turns or corners at the points where the expression inside the absolute value reaches zero. These points, frequently called creases, influence the differentiation of the function.
The function \( x^3 - 9x \) within \( f(x) \) is zero at points like \(x = 0, 3, -3\). As a result, the absolute value introduces non-differentiable corners at these points, affecting the nature of critical points found.
The function \( x^3 - 9x \) within \( f(x) \) is zero at points like \(x = 0, 3, -3\). As a result, the absolute value introduces non-differentiable corners at these points, affecting the nature of critical points found.
Critical Points
Critical points are locations on a function's graph where the derivative is zero or doesn't exist, potentially indicating extremum or other significant behavior changes. For \( f(x) = |x^3 - 9x| \), we solve \( x^3 - 9x = 0 \) to get the critical points of the function, giving values \(x = 0, 3, -3\).
At these points, the absolute value implies that we must carefully check for differentiability. While \( f'(x) \) might exist at \( x = 0 \) (giving \( f'(0) = -9 \)), the corners at \( x = 3 \) and \( x = -3 \) mean \( f'(3) \) and \( f'(-3) \) do not exist. Critical points thus help locate potential extrema, as further analysis of interval signs is needed to confirm extremum status.
At these points, the absolute value implies that we must carefully check for differentiability. While \( f'(x) \) might exist at \( x = 0 \) (giving \( f'(0) = -9 \)), the corners at \( x = 3 \) and \( x = -3 \) mean \( f'(3) \) and \( f'(-3) \) do not exist. Critical points thus help locate potential extrema, as further analysis of interval signs is needed to confirm extremum status.
Derivative Tests
Derivative tests are powerful tools in calculus for understanding the behavior of a function, particularly around critical points. These tests use the derivative \( f'(x) \) to determine the nature of critical points. The first derivative test involves checking the sign of \( f'(x) \) before and after a critical point to see if it shifts from positive to negative or vice versa.
In the exercise with \( f(x) = |x^3 - 9x| \), the derivative \( f'(x) = 3x^2 - 9 \) is used to evaluate the critical points. By examining intervals around these points, we noted the behavior of the function. For example, at \(x = -3\) and \(x = 3\), transitions from negative to positive sign indicated local minima due to the absolute value's influence. Derivative tests are essential for understanding these finer details.
In the exercise with \( f(x) = |x^3 - 9x| \), the derivative \( f'(x) = 3x^2 - 9 \) is used to evaluate the critical points. By examining intervals around these points, we noted the behavior of the function. For example, at \(x = -3\) and \(x = 3\), transitions from negative to positive sign indicated local minima due to the absolute value's influence. Derivative tests are essential for understanding these finer details.
Other exercises in this chapter
Problem 77
Find the a$$0$$bsolute maximum and minimum values of \(f(x)=\) \(e^{x}-2 x\) on [0,1]
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Only one of these calculations is correct. Which one? Why are the others wrong? Give reasons for your answers. a. \(\lim _{x \rightarrow 0^{+}} x \ln x=0 \cdot(
View solution Problem 78
Verify the formulas in Exercises by differentiation. $$\int x e^{x} d x=x e^{x}-e^{x}+C$$
View solution Problem 78
Show that \(\left(e^{x_{1}}\right)^{x_{2}}=e^{x_{1} x_{2}}=\left(e^{x_{2}}\right)^{x_{1}}\) for any numbers \(x_{1}\) and \(x_{2}\).
View solution