Question

Sketch the graph of \(g(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}}, & x \neq 0 \\ 0, & x=0\end{array}\right.\) and determine \(g^{\prime}(0)\).

Short answer

The sketched graph of the function would display a horizontal asymptote at y=0, with the graph lying above the x-axis but never touching or crossing it. The derivative of g at x=0 is 0.

Step-by-step solution

Analyze the function

Understand the nature of the given function. The function takes two distinct forms: \(e^{-1 / x^{2}}\) for x ≠ 0 and 0 for x = 0.

Sketch the function.

For x ≠ 0, the function \(e^{-1 / x^{2}}\) will always give positive values close to zero since the exponent is a negative reciprocal. As x gets close to 0 from both sides, the output of the function tends towards 0, which agrees with the definition of the function at x = 0. So, sketch a graph that displays this behavior, with a horizontal asymptote at y=0.

Find the derivative of \(e^{-1 / x^{2}}\)

Firstly, let's recognize that we are dealing with the function of a function here (Chain rule applies). So, differentiate \(e^{-1 / x^{2}}\) with respect to x, by applying the chain rule. The chain rule dictates that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function. Therefore the derivative of \(e^{-1 / x^{2}}\) is \(-2e^{-1 / x^{2}}/x^{3} = -2g(x)/x^{3}\) for \(x \neq 0\). Note: this result is not valid for x=0.

Determine the derivative at x=0

To derive the function at x=0, we need to verify if the limit \( \lim_{{x \to 0}} {g^{'}(x)}\) exists. The function g(x) has the same value, 0, when approached from both sides of x=0, therefore the limit exists. This implies that g'(0) = 0.