Problem 7
Question
2-7. Let \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) be a function such that \(|f(x)| \leq|x|^{2}\). Show that \(f\) is differentiable at 0 .
Step-by-Step Solution
Verified Answer
Thus, \( f \) is differentiable at 0.
1Step 1 - Definition of Differentiability
To show that a function is differentiable at a point, we need to show that the limit \[ \lim_{{h \to 0}} \frac{{f(h) - f(0) - abla f(0) \cdot h}}{|h|} = 0 \]holds.
2Step 2 - Evaluate at Zero
First, evaluate the function at zero. Since \( f: \mathbf{R}^n \rightarrow \mathbf{R} \) and \(|f(x)| \leq |x|^2\), we have \( f(0) = 0 \).
3Step 3 - Boundary Behavior
Given that \(|f(x)| \leq |x|^2\), we can use this to bound \(f(h)\) as \(|f(h)| \leq |h|^2\).
4Step 4 - Consider the Differentiability Limit
To check the differentiability limit, notice that\[ \left| \frac{{f(h) - f(0) - abla f(0) \cdot h}}{|h|} \right| = \left| \frac{f(h)}{|h|} \right| \].By the bound, we have \( \left| \frac{f(h)}{|h|} \right| \leq \frac{|h|^2}{|h|} = |h| \).
5Step 5 - Taking the Limit
Taking the limit as \( h \to 0 \), it is clear that \( |h| \to 0 \). Thus,\[ \lim_{{h \to 0}} \left| \frac{f(h)}{|h|} \right| = 0 \].
Key Concepts
Limit Definition of DifferentiabilityBounding FunctionsBehavior at ZeroEpsilon-Delta Definition of Differentiability
Limit Definition of Differentiability
Differentiability at a point ensures that the function behaves nicely around that point. According to the limit definition of differentiability, we need to show that the limit: \[ \boxed{ \text{lim}_{{h \to 0}} \frac{{f(h) - f(0) - abla f(0) \bullet h}}{|h|} = 0 } \]holds. Here, 𝑓(ℎ) represents the function at point ℎ, 𝑓(0) is the function at zero, and ∇𝑓(0)·h is the gradient of the function at zero multiplied by ℎ. A function being differentiable at a specific point means:
- The derivative exists at that point.
- The function is smooth, without any abrupt changes, at that point.
Bounding Functions
In this problem, bounding the function 𝑓(𝑥) is essential. We know that the function satisfies the inequality: \[ |f(x)| \boxed{\text{≤}} |x|^{2} \] This tells us that the absolute value of the function is always less than or equal to the square of 𝑥. Bounding functions helps in:
- Understanding the maximum possible values the function can take.
- Providing useful estimates to solve limits and integrals.
Behavior at Zero
To show differentiability at zero, it is important to evaluate the function at zero. From the given bound \(|f(x)| \boxed{\text{≤}} |x|^{2} \), we get that: \( f(0) \boxed{= 0} \) The behavior of the function at zero is important because it serves as a reference point when we analyze small changes around zero. Knowing the value of the function at zero, we can better understand how the function behaves as it approaches zero. Understanding the behavior at zero leads us to more easily test conditions for differentiability, as shown in this problem. It provides a starting point for our limit calculations.
Epsilon-Delta Definition of Differentiability
Another way to test differentiability is using the Epsilon-Delta definition. This definition states that a function 𝑓 is differentiable at a point a if:
- For every ϵ > 0, there exists a δ > 0 such that for all h,
- if |h| < δ, then
- \( \frac{|f(h) - f(0) - ∇f(0)·h|}{|h|} \boxed{\text{<}} ϵ \).
- We want the difference between the actual function value and the linear approximation at zero to be less than ϵ,
- whenever ℎ is within a specific range defined by δ.
Other exercises in this chapter
Problem 4
2-4. Let \(g\) be a continuous real-valued function on the unit circle \(\left\\{x \in \mathbf{R}^{2}:|x|=1\right\\}\) such that \(g(0,1)=g(1,0)=0\) and \(g(-x)
View solution Problem 6
2-6. Let \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be defined by \(f(x, y)=\sqrt{|x y|}\). Show that \(f\) is not differentiable at \((0,0)\).
View solution Problem 8
2-8. Let \(f: \mathbf{R} \rightarrow \mathbf{R}^{2}\). Prove that \(f\) is differentiable at \(a \in \mathbf{R}\) if and only if \(f^{1}\) and \(f^{2}\) are, an
View solution Problem 9
2-9. Two functions \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) are equal up to \(\boldsymbol{n}\) th order at \(a\) if $$ \lim _{h \rightarrow 0} \frac{f(a+h)-g
View solution