Let's analyze the given problem step by step. --- ### **Given Function:** \[ f: \mathbb{R} \to \mathbb{R} \] \[ f(x) = 11x^2 + 112x + e^2 \] *Note:* The original description seems to have some formatting issues, but based on typical forms, I'll interpret the function as: \[ f(x) = 11x^2 + 112x + e^2 \] --- ### **Step 1: Find the Gradient \( \nabla f(x) \)** Since \(f:\mathbb{R} \to \mathbb{R}\), the gradient reduces to the derivative: \[ f'(x) = \frac{d}{dx} (11x^2 + 112x + e^2) \] Calculate term by term: - \(\frac{d}{dx} (11x^2) = 22x\) - \(\frac{d}{dx} (112x) = 112\) - \(\frac{d}{dx} (e^2) = 0\) (since \(e^2\) is a constant) **Therefore:** \[ \boxed{ \nabla f(x) = f'(x) = 22x + 112 } \] --- ### **Step 2: Show that \(f\) is Frechet differentiable on \(\mathbb{R}\)** In finite-dimensional real spaces, the Frechet derivative coincides with the usual derivative for functions that are continuously differentiable. **Recall:** - A function \(f:\mathbb{R} \to \mathbb{R}\) is Frechet differentiable at \(x\) if there exists a bounded linear operator \(A:\mathbb{R} \to \mathbb{R}\) such that: \[ \lim_{h \to 0} \frac{|f(x+h) - f(x) - A(h)|}{|h|} = 0 \] - When \(f\) is differentiable at \(x\), the Frechet derivative \(A\) is given by: \[ A(h) = f'(x) \cdot h \] --- ### **Step 3: Verify the differentiability** Since \(f\) is a polynomial plus an exponential constant term, it is continuous and differentiable everywhere. - The derivative \(f'(x) = 22x + 112\) exists for all \(x \in \mathbb{R}\) and is continuous. - Polynomial and exponential functions are smooth functions, so their derivatives are continuous everywhere. **Explicitly:** \[ f(x+h) = 11(x+h)^2 + 112(x+h) + e^2 \] \[ = 11(x^2 + 2xh + h^2) + 112x + 112h + e^2 \] \[ = 11x^2 + 22x h + 11 h^2 + 112x + 112 h + e^2 \] Compare \(f(x+h)\) and \(f(x)\): \[ f(x+h) - f(x) = (11x^2 + 22x h + 11 h^2 + 112x + 112 h + e^2) - (11x^2 + 112x + e^2) = 22x h + 11 h^2 + 112 h \] Factor \(h\): \[ f(x+h) - f(x) = h (22x + 112) + 11 h^2 \] Define \(A(h) = f'(x) \cdot h = (22x + 112) h\). Then, \[ f(x+h) - f(x) - A(h) = 11 h^2 \] Now, compute the limit: \[ \frac{|f(x+h) - f(x) - A(h)|}{|h|} = \frac{|11 h^2|}{|h|} = 11 |h| \to 0 \quad \text{as} \quad h \to 0 \] since \(h \to 0\), the expression tends to zero. --- ### **### **Conclusion:**** - **Gradient:** The gradient (or derivative) of \(f\) at any point \(x\) is: \[ \boxed{ \nabla f(x) = 22x + 112 } \] - **Frechet Differentiability:** The function \(f(x) = 11x^2 + 112x + e^2\) is **Frechet differentiable everywhere on \(\mathbb{R}\)**, with derivative \(A(h) = (22x + 112)h\). --- ### **Summary:** | Step | Result | Explanation | |---------|------------------------|------------------------------| | 1 | \(\nabla f(x) = 22x + 112\) | Derivative of a quadratic polynomial plus constant | | 2 | \(f\) is Frechet differentiable on \(\mathbb{R}\) | Polynomial and exponential functions are smooth, making \(f\) infinitely differentiable | --- **Feel free to ask if you'd like further clarification or additional details!**

Let’s carefully solve the following problem step by step: --- ## **Given:** Let \( f : \mathbb{R} \to \mathbb{R} \) be given by \[ f(x) = \|x\|_2^2 \cdot e^{-\|x\|} \] where \(\|x\|_2\) is the usual Euclidean norm. **But since \(x \in \mathbb{R}\) (i.e., \(x\) is a real number, not a vector), \(\|x\|_2 = |x|\).** So, \[ f(x) = |x|^2 \cdot e^{-|x|} = x^2 e^{-|x|} \] --- ## **Step 1: Compute the Gradient (Derivative) \( \nabla f(x) \)** Since \(f:\mathbb{R} \to \mathbb{R}\), the gradient is just the derivative. Let’s compute \(f'(x)\): \[ f(x) = x^2 e^{-|x|} \] We need to be careful with the absolute value in the exponent. Consider two cases: \(x \geq 0\) and \(x < 0\). --- ### **Case 1: \(x \geq 0\)** Then \(|x| = x\), so \[ f(x) = x^2 e^{-x} \] Now, \[ f'(x) = \frac{d}{dx} (x^2 e^{-x}) \] Using the product rule: \[ = 2x e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} + x^2 (-e^{-x}) = 2x e^{-x} - x^2 e^{-x} = (2x - x^2) e^{-x} \] --- ### **Case 2: \(x < 0\)** Then \(|x| = -x\), so \[ f(x) = x^2 e^{-(-x)} = x^2 e^{x} \] Now, \[ f'(x) = \frac{d}{dx} (x^2 e^{x}) \] Again, product rule: \[ = 2x e^{x} + x^2 e^{x} = (2x + x^2) e^{x} \] --- ### **Combine the Cases** So, the derivative is: \[ f'(x) = \begin{cases} (2x - x^2) e^{-x} & \text{if } x \geq 0 \\ (2x + x^2) e^{x} & \text{if } x < 0 \end{cases} \] Or, more compactly, using the sign function \(\operatorname{sgn}(x)\): \[ f'(x) = \left(2x - x^2 \cdot \operatorname{sgn}(x)\right) e^{-|x|} \] But the piecewise form is often clearer. --- ## **Step 2: Show That \(f\) is Fréchet Differentiable on \(\mathbb{R}\)** For functions from \(\mathbb{R}\) to \(\mathbb{R}\), Fréchet differentiability at a point \(x_0\) means: There exists a linear map \(A(h)\) such that: \[ \lim_{h \to 0} \frac{|f(x_0 + h) - f(x_0) - A(h)|}{|h|} = 0 \] For real functions, \(A(h) = f'(x_0) h\). ### **Check Differentiability at \(x \neq 0\):** - Both \(f(x)\) and \(f'(x)\) are clearly continuous and differentiable for \(x \neq 0\) (since exponentials and polynomials are smooth). ### **Check Differentiability at \(x = 0\):** - For \(x=0\), check if the derivative exists and is continuous. Compute left and right derivatives: #### **Right derivative (\(x \to 0^+\)):** \[ f(x) = x^2 e^{-x},\ f'(x) = (2x - x^2) e^{-x} \] So, \[ f'(0^+) = (2 \cdot 0 - 0^2) e^{0} = 0 \] #### **Left derivative (\(x \to 0^-\)):** \[ f(x) = x^2 e^{x},\ f'(x) = (2x + x^2) e^{x} \] So, \[ f'(0^-) = (2 \cdot 0 + 0^2) e^{0} = 0 \] So the derivative exists and is 0 at \(x=0\). #### **Is the derivative continuous at \(x=0\)?** - As \(x \to 0\), both \(f'(x)\) from the left and right approach 0. - Therefore, \(f'\) is continuous at \(x=0\). ### **Conclusion on Fréchet Differentiability:** - Since \(f\) is differentiable everywhere and \(f'\) is continuous everywhere, \(f\) is Fréchet differentiable everywhere on \(\mathbb{R}\). --- ## **Final Answers** ### **1. Gradient (derivative) at any point \(x\):** \[ \boxed{ f'(x) = \begin{cases} (2x - x^2) e^{-x} & \text{if } x \geq 0 \\ (2x + x^2) e^{x} & \text{if } x < 0 \end{cases} } \] ### **2. Fréchet differentiability:** \[ \boxed{ f(x) = x^2 e^{-|x|} \text{ is Fréchet differentiable everywhere on } \mathbb{R} } \] --- ## **Summary Table** | Step | Result | Explanation | |------|--------|-------------| | 1 | \(f'(x)\) as above | Computed using product and chain rule, case by case | | 2 | \(f\) Fréchet differentiable on \(\mathbb{R}\) | \(f\) is differentiable everywhere, and the derivative is continuous | --- **Let me know if you’d like to see the multivariable version or further generalization!**