Problem 28
Question
In Exercises 28 through 31 , find formulas for \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) and state the domains of \(f^{\prime}\) and \(f^{\prime \prime}\). $$ f(x)= \begin{cases}\frac{x^{2}}{|x|} & \text { if } x \neq 0 \\ 0 & \text { if } x=0\end{cases} $$
Step-by-Step Solution
Verified Answer
For \(x>0\), \(f'(x)=1\), \(f''(x)=0\). For \(x<0\), \(f'(x)=-1\), \(f''(x)=0\). The domains of \(f'\) and \(f''\) are \(x eq 0\).
1Step 1: Rewrite the function
Rewrite the function for different cases: If \(x>0\), \(f(x) = x\) If \(x<0\), \(f(x) = -x\) If \(x=0\), \(f(x)=0\)
2Step 2: Find the first derivative for each case
Compute the derivative of \(f(x)\) for each interval: For \(x>0\), \(f'(x) = 1\) For \(x<0\), \(f'(x) = -1\).
3Step 3: Find the second derivative for each case
Since the first derivatives are constants, compute the second derivative: For \(x>0\), \(f''(x) = 0\) For \(x<0\), \(f''(x) = 0\).
4Step 4: State the domains of \(f'(x)\) and \(f''(x)\)
The domain of \(f'(x)\) and \(f''(x)\) is \(x eq 0\)
Key Concepts
First DerivativeSecond DerivativePiecewise FunctionCalculus
First Derivative
The first derivative, denoted as \( f'(x) \), represents the rate at which a function changes at any given point. In other words, it gives the slope of the tangent line to the function at any point on its curve. For the provided piecewise function, we need to analyze the behavior of the function in different intervals: when \( x > 0 \) and when \( x < 0 \), and at \( x = 0 \).
For \( x > 0 \), the function behaves linearly as \( f(x) = x \). The derivative of \( x \) is simply \( 1 \) because the slope of the line \( y = x \) is constant and equals 1 everywhere.
For \( x < 0 \), the function behaves as \( f(x) = -x \). The derivative of \( -x \) is \( -1 \) because the slope is constantly -1 for negative \( x \)-values.
Therefore, the derivative for \( x > 0 \) is \( f'(x) = 1 \) and for \( x < 0 \), it is \( f'(x) = -1 \). At \( x = 0 \), the function is not differentiable because the left-hand and right-hand limits do not match.
For \( x > 0 \), the function behaves linearly as \( f(x) = x \). The derivative of \( x \) is simply \( 1 \) because the slope of the line \( y = x \) is constant and equals 1 everywhere.
For \( x < 0 \), the function behaves as \( f(x) = -x \). The derivative of \( -x \) is \( -1 \) because the slope is constantly -1 for negative \( x \)-values.
Therefore, the derivative for \( x > 0 \) is \( f'(x) = 1 \) and for \( x < 0 \), it is \( f'(x) = -1 \). At \( x = 0 \), the function is not differentiable because the left-hand and right-hand limits do not match.
Second Derivative
The second derivative, denoted as \( f''(x) \), measures the concavity of the function or the rate at which the first derivative changes. In simpler terms, it indicates how the slope of the function itself is varying. For the function provided, let us again consider the intervals separately.
Since the first derivatives \( f'(x) \) are constant in the intervals \( x > 0 \) and \( x < 0 \), their second derivatives will be zero. This is because the derivative of a constant is always zero.
This leads us to:
There is no second derivative at \( x = 0 \) because the function is not differentiable at that point.
Since the first derivatives \( f'(x) \) are constant in the intervals \( x > 0 \) and \( x < 0 \), their second derivatives will be zero. This is because the derivative of a constant is always zero.
This leads us to:
- For \( x > 0 \): \( f''(x) = 0 \)
- For \( x < 0 \): \( f''(x) = 0 \)
There is no second derivative at \( x = 0 \) because the function is not differentiable at that point.
Piecewise Function
A piecewise function is a function that is defined by different expressions in different parts of its domain. These different pieces can follow their own rules and equations, creating a combined function. For example, the function given in the exercise is a classic piecewise function. It is defined as \( f(x) = \frac{x^2}{|x|} \) if \( x eq 0 \) and as \( f(x) = 0 \) if \( x = 0 \).
To solve problems involving piecewise functions, remember to:
Always be careful when considering the domain of each piece and ensure a unified understanding of the whole function.
To solve problems involving piecewise functions, remember to:
- Identify each piece and the intervals they apply to.
- Analyze the behavior of the function within each interval separately.
- Ensure the function and its derivatives match up at boundary points, if necessary.
Always be careful when considering the domain of each piece and ensure a unified understanding of the whole function.
Calculus
Calculus is a branch of mathematics that studies continuous change. The two primary operations in calculus are differentiation and integration. Differentiation is concerned with finding derivatives, which represent rates of change. Integration is the reverse process of differentiation and is concerned with finding areas under curves.
Calculus is incredibly powerful for understanding real-world phenomena where things are changing continuously. Key concepts of calculus include:
In this context, understanding how to work with piecewise functions and finding their derivatives is crucial. More advanced applications include solving differential equations and optimizing functions.
Calculus is incredibly powerful for understanding real-world phenomena where things are changing continuously. Key concepts of calculus include:
- Limits: Fundamental for understanding continuity and for defining derivatives and integrals.
- Derivatives: Measures how a function changes as its input changes, denoted as \( f'(x) \).
- Integrals: Measures the accumulation of quantities, such as areas under curves.
In this context, understanding how to work with piecewise functions and finding their derivatives is crucial. More advanced applications include solving differential equations and optimizing functions.
Other exercises in this chapter
Problem 27
Find the derivative of the given function. (HIN\Upsilon: \(\left.|a|=\sqrt{a^{2}} .\right)\) $$ g(x)=|x|^{3} $$
View solution Problem 27
An equation is given. Do the following in each of these problems: (a) Find two functions defined by the equation, and state their domains. (b) Draw a sketch of
View solution Problem 28
Use the chain rule to prove that (a) the derivative of an even function is an odd function, and (b) the derivative of an odd function is an even function, provi
View solution Problem 28
Find the derivative of the given function. (HIN\Upsilon: \(\left.|a|=\sqrt{a^{2}} .\right)\) $$ h(x)=\sqrt[3]{|x|+x} $$
View solution