Understanding Differentiability of Functions

The question asks for the set of all points where the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable. To determine where a function is differentiable, we first need to understand its domain and then evaluate its derivative.

A function \(f(x)\) is differentiable at a point \(a\) if the limit \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \) exists. For functions involving square roots like \(f(x) = \sqrt{g(x)}\), differentiability depends on two main conditions:

• The function \(g(x)\) inside the square root must be differentiable.
• The value of \(g(x)\) must be positive, i.e., \(g(x) > 0\), at the point of differentiability. If \(g(x) = 0\), the derivative of \(\sqrt{g(x)}\) might be undefined due to division by zero in the derivative formula.

Step-by-Step Analysis of the Function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \)

1. Find the Domain of the Function \({\rm{f}}\left( {\rm{x}} \right)\)

For \({\rm{f}}\left( {\rm{x}} \right)\) to be defined, the expression under the square root must be non-negative.

We need \(1 - {{\rm{e}}^{ - {{\rm{x}}^2}}} \ge 0\).

This inequality can be rewritten as \(1 \ge {{\rm{e}}^{ - {{\rm{x}}^2}}}\).

Since the natural logarithm function \(\ln(y)\) is an increasing function, we can take the natural logarithm of both sides without changing the inequality direction:

\(\ln(1) \ge \ln({{\rm{e}}^{ - {{\rm{x}}^2}}})\)

\(0 \ge -{{\rm{x}}^2}\)

Multiplying by -1 and reversing the inequality sign, we get:

\({{\rm{x}}^2} \ge 0\)

This inequality \(x^2 \ge 0\) is true for all real numbers \(x\). Therefore, the domain of the function \({\rm{f}}\left( {\rm{x}} \right)\) is \((-\infty, \infty)\).

2. Find the Derivative of the Function \({\rm{f}}\left( {\rm{x}} \right)\)

To find where the function is differentiable, we need to compute its derivative using the chain rule. Let \(u = 1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}\). Then \(f(x) = \sqrt{u}\). The derivative of \(\sqrt{u}\) with respect to \(u\) is \(\frac{1}{2\sqrt{u}}\). We also need to find the derivative of \(u = 1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}\) with respect to \(x\).

Let \(v = -x^2\). Then \(u = 1 - e^v\). The derivative of \(1 - e^v\) with respect to \(v\) is \(-e^v\). The derivative of \(v = -x^2\) with respect to \(x\) is \(-2x\).

Using the chain rule: \(\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = (-e^v) \cdot (-2x) = (-e^{-x^2}) \cdot (-2x) = 2x{{\rm{e}}^{ - {{\rm{x}}^2}}}\).

Now, applying the chain rule to \(f(x) = \sqrt{u}\):

\({{\rm{f}}'}\left( {\rm{x}} \right) = \frac{d}{du}(\sqrt{u}) \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot (2x{{\rm{e}}^{ - {{\rm{x}}^2}}})\)

Substitute \(u = 1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}\) back:

\({{\rm{f}}'}\left( {\rm{x}} \right) = \frac{1}{2\sqrt{1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}}} \cdot (2x{{\rm{e}}^{ - {{\rm{x}}^2}}})\)

\({{\rm{f}}'}\left( {\rm{x}} \right) = \frac{x{{\rm{e}}^{ - {{\rm{x}}^2}}}}{\sqrt{1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}}}\)

3. Determine Where the Derivative Exists

For the derivative \({{\rm{f}}'}\left( {\rm{x}} \right) = \frac{x{{\rm{e}}^{ - {{\rm{x}}^2}}}}{\sqrt{1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}}}\) to exist, the following conditions must be met:

• The expression in the numerator \(x{{\rm{e}}^{ - {{\rm{x}}^2}}}\) must be defined. Since \(x\) is a real number and \(e^{-x^2}\) is defined for all real \(x\), the numerator is defined for all \(x \in (-\infty, \infty)\).
• The expression in the denominator \(\sqrt{1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}}\) must be defined and non-zero.
  • It is defined when \(1 - {{\rm{e}}^{ - {{\rm{x}}^2}}} \ge 0\), which we already found is true for all \(x \in (-\infty, \infty)\).
  • It must be non-zero, which means \(\sqrt{1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \ne 0\). This requires \(1 - {{\rm{e}}^{ - {{\rm{x}}^2}}} \ne 0\).

We need \(1 - {{\rm{e}}^{ - {{\rm{x}}^2}}} \ne 0\).

\(1 \ne {{\rm{e}}^{ - {{\rm{x}}^2}}}\)

\(\ln(1) \ne \ln({{\rm{e}}^{ - {{\rm{x}}^2}}})\)

\(0 \ne -{{\rm{x}}^2}\)

\({{\rm{x}}^2} \ne 0\)

This inequality holds true for all real numbers \(x\) except for \(x=0\).

So, the derivative exists for all \(x \in (-\infty, \infty)\) such that \(x \ne 0\). This set is \((-\infty, 0) \cup (0, \infty)\).

At \(x=0\), the function \(f(0) = \sqrt{1 - e^{-0^2}} = \sqrt{1 - e^0} = \sqrt{1 - 1} = \sqrt{0} = 0\). The derivative formula involves \(\sqrt{1 - e^{-x^2}}\) in the denominator, which is 0 at \(x=0\), making the formula undefined. We would need to check the definition of the derivative at \(x=0\), but the term \(\sqrt{g(x)}\) typically fails to be differentiable at points where \(g(x)=0\), unless \(g'(x)=0\) as well (which is not the case here, as \(g'(0) = 2(0)e^{-0^2} = 0\), so the limit \(\frac{f(x)-f(0)}{x-0} = \frac{\sqrt{1-e^{-x^2}}}{x}\) as \(x \to 0\) needs careful evaluation, likely showing it does not exist). The standard rule for \(\sqrt{g(x)}\) is differentiability requires \(g(x) > 0\).

Conclusion on Differentiability

Based on the analysis of the derivative formula and the conditions for differentiability of a square root function, the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable at all points \(x\) where \(1 - {{\rm{e}}^{ - {{\rm{x}}^2}}} > 0\). This occurs when \(x^2 > 0\), which means \(x \ne 0\).

Therefore, the set of all points where the function is differentiable is \((-\infty, 0) \cup (0, \infty)\).

Revision Table: Key Concepts for Differentiability

Additional Information: Exploring Related Concepts

The function \(e^{-x^2}\) is a fundamental part of the Gaussian function, widely used in probability and statistics. The expression \(1 - e^{-x^2}\) ranges from 0 (at \(x=0\)) up to 1 (as \(|x| \to \infty\)).

The non-differentiability at \(x=0\) for functions like \(\sqrt{g(x)}\) when \(g(0)=0\) typically occurs because the graph might have a "sharp turn" or a vertical tangent at that point. For \(f(x) = \sqrt{1 - e^{-x^2}}\), let's consider the behavior near \(x=0\).

Using the Taylor series expansion for \(e^u\) around \(u=0\), \(e^u \approx 1 + u + \frac{u^2}{2}\). Let \(u = -x^2\). As \(x \to 0\), \(u \to 0\).

\(e^{-x^2} \approx 1 + (-x^2) + \frac{(-x^2)^2}{2} = 1 - x^2 + \frac{x^4}{2}\)

So, \(1 - e^{-x^2} \approx 1 - (1 - x^2 + \frac{x^4}{2}) = x^2 - \frac{x^4}{2}\) for small \(x\).

Then \(f(x) = \sqrt{1 - e^{-x^2}} \approx \sqrt{x^2 - \frac{x^4}{2}} = \sqrt{x^2(1 - \frac{x^2}{2})} = |x|\sqrt{1 - \frac{x^2}{2}}\).

As \(x \to 0\), \(\sqrt{1 - \frac{x^2}{2}} \to \sqrt{1} = 1\). So \(f(x) \approx |x|\) near \(x=0\). The function \(|x|\) is known to be not differentiable at \(x=0\) due to a sharp point. This heuristic supports the analytical finding that \(f(x)\) is not differentiable at \(x=0\).

The set of all points where the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable is \((-\infty, 0) \cup (0, \infty)\).