Problem 4
Question
Define \(f: \mathbb{R} \rightarrow \mathbb{R}\) by $$ f(x)=\left\\{\begin{array}{ll} x^{2}, & \text { if } x<0 \\ x^{3}, & \text { if } x \geq 0 \end{array}\right. $$ Is \(f\) differentiable at \(0 ?\)
Step-by-Step Solution
Verified Answer
Yes, \(f\) is differentiable at 0.
1Step 1: Definition of Differentiability
A function \(f\) is differentiable at a point \(a\) if the limit \(\lim_{x \to a}\frac{f(x)-f(a)}{x-a}\) exists.
2Step 2: Evaluate \(f(0)\)
Calculate \(f(0)\) using the piecewise definition of \(f\). Since \(x = 0\), we use \(f(x) = x^3\). Thus, \(f(0) = 0^3 = 0\).
3Step 3: Calculate the Left-Hand Limit
For \(x < 0\), \(f(x) = x^2\). Compute the limit \(\lim_{x \to 0^-}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^-}\frac{x^2}{x} = \lim_{x \to 0^-} x = 0\).
4Step 4: Calculate the Right-Hand Limit
For \(x \geq 0\), \(f(x) = x^3\). Compute the limit \(\lim_{x \to 0^+}\frac{f(x)-f(0)}{x-0} = \lim_{x \to 0^+}\frac{x^3}{x} = \lim_{x \to 0^+} x^2 = 0\).
5Step 5: Compare the Limits
The left-hand limit and right-hand limit at \(x=0\) are both 0. Therefore, \(\lim_{x \to 0}\frac{f(x)-f(0)}{x-0} = 0\) exists.
6Step 6: Conclusion on Differentiability
Since both the left-hand and right-hand limits exist and are equal, \(f\) is differentiable at \(x=0\).
Key Concepts
Piecewise functionsLeft-hand limitRight-hand limit
Piecewise functions
A piecewise function is a type of function that is defined by different expressions depending on the input value. Each "piece" or part of the function applies to a certain interval of the domain. This can be helpful in modeling real-world situations where the behavior of a function changes at different intervals.
- For example, the function given in the original exercise is defined as \( f(x) = x^2 \) when \( x < 0 \), and as \( f(x) = x^3 \) when \( x \geq 0 \).
- Such functions are often used to model situations where a different rule needs to be applied depending on certain conditions.
Left-hand limit
The left-hand limit refers to the value that a function approaches as the input approaches a particular point from the left-hand side. This means that we're considering values just below the point in question.
- Mathematically, for a point \( a \), the left-hand limit is denoted as \( \lim_{x \to a^-} f(x) \).
- In the context of the original exercise for the function \( f \) at \( x = 0 \), we examined this limit by looking at the segment of the piecewise function where \( f(x) = x^2 \) and calculating \( \lim_{x \to 0^-} \frac{x^2}{x} = \lim_{x \to 0^-} x = 0 \).
Right-hand limit
The right-hand limit is analogous to the left-hand limit, but it considers the approach from the right-hand side of a point. This involves analyzing the function values that are just above the point.
- This limit can be expressed mathematically as \( \lim_{x \to a^+} f(x) \) for a given point \( a \).
- In our example function at \( x = 0 \), the right-hand limit is evaluated for the piece where \( f(x) = x^3 \), leading to \( \lim_{x \to 0^+} \frac{x^3}{x} = \lim_{x \to 0^+} x^2 = 0 \).
Other exercises in this chapter
Problem 4
Suppose \(f \in C^{(2)}(a, b)\). Use Taylor's theorem to show that $$ \lim _{h \rightarrow 0} \frac{f(c+h)+f(c-h)-2 f(c)}{h^{2}}=f^{\prime \prime}(c) $$ for any
View solution Problem 4
Prove the Mean Value Theorem using Rolle's theorem and the function $$ k(t)=f(t)-\left(\left(\frac{f(b)-f(a)}{b-a}\right)(t-a)+f(a)\right) . $$ Give a geometric
View solution Problem 5
Suppose \(f \in C^{(1)}(a, b), c \in(a, b), f^{\prime}(c)=0,\) and \(f^{\prime \prime}\) exists on \((a, b)\) and is continuous at \(c .\) Show that \(f\) has a
View solution Problem 5
Let \(a, b \in \mathbb{R}\). Suppose \(f\) is continuous on \([a, b]\), differentiable on \((a, b),\) and \(\left|f^{\prime}(x)\right| \leq M\) for all \(x \in(
View solution