91Ó°ÊÓ

The domain of a function is the set of all possible input values (x-values) for which the function is defined. Since the exponent in the function is a real number (namely, \(-x^2/2\)), there are no restrictions on the domain. Therefore, the domain of the function is: Domain: \((-\infty, +\infty)\)

The range of a function is the set of all possible output values (y-values) produced by the function. Since the base of the exponential function is the number "e" (which is always positive) and the exponent will never be positive (\(-x^2/2\) is always non-positive), the output values of the function will always be between 0 and 1 (excluding 0). Thus, the range of the function is: Range: \((0, 1]\)

To determine the increasing and decreasing intervals, we need to first find the derivative of the function. $$f'(x)=\frac{d}{dx}(e^{-x^2/2})$$ Using the chain rule, we get: $$f'(x)=e^{-x^2/2}\cdot(-x)= -xe^{-x^2/2}$$ Now, we need to find the critical points where the derivative is equal to 0 or undefined. In this case, the derivative is never undefined (since it is an exponential function). So, we only need to find where \(f'(x)=0\). Since \(e^{-x^2/2}\) is always positive, we can focus on the other part: \(-x=0\). The only critical point is at \(x=0\). Now, we can determine the intervals where the function is increasing and decreasing: Increasing interval: \((-\infty, 0)\) Decreasing interval: \((0, +\infty)\)

Since there is only one critical point (\(x=0\)), and we know the function is increasing before the critical point and decreasing after, we can conclude that there is a local maximum at \(x=0\). To find the y-value of this local maximum, we can plug \(x=0\) into the original function \(f(x)\): \(f(0)=e^{-0^2/2}=e^0=1\) Thus, there is a local maximum at the point \((0,1)\).

Since the function is an exponential function, there is a horizontal asymptote at y=0. There are no vertical asymptotes. To determine the concavity, we need to find the second derivative: $$f''(x)=\frac{d^2}{dx^2}\left(-xe^{-x^2/2}\right)$$ Applying the product rule, we get: $$f''(x)=-e^{-x^2/2}+x^2 e^{-x^2/2}$$ We identify the inflection points (where concavity changes) by finding where the second derivative is equal to 0 or undefined. Again, we only need to look at where it is equal to 0. Notice that the exponential term is always positive, so we only need to look at the other terms: $$0=-1+x^2\implies x=\pm\sqrt{1}\implies x=\pm1$$ We have two inflection points at \(x=-1\) and \(x=1\). Using these points, we can determine the concavity intervals: Concave up: \((-\infty, -1)\cup(1, \infty)\) Concave down: \((-1, 1)\)

Using all the information we gathered in the previous steps, the graph of the function \(f(x)=e^{-x^2/2}\) will have the following features: 1. Domain: \((-\infty, +\infty)\) 2. Range: \((0, 1]\) 3. Increasing interval: \((-\infty, 0)\) 4. Decreasing interval: \((0, +\infty)\) 5. Local maximum at (0, 1) 6. Horizontal asymptote: y=0 7. Concave up: \((-\infty, -1)\cup(1, \infty)\) 8. Concave down: \((-1, 1)\) Now, you can draw the complete graph of the function \(f(x)=e^{-x^2/2}\) using the gathered information.