Mark the critical points on the following graph. \( x^{7} e^{-\frac{x^{2}}{4}}, 62 \)

We start with the function \[ f(x)=x^7e^{-\frac{x^2}{4}}. \] A critical point occurs when the derivative \( f'(x) \) is zero or undefined. Since \( f(x) \) is smooth everywhere, we only need to solve \( f'(x)=0 \). ### Step 1. Compute \( f'(x) \) Using the product rule for differentiation, we have \[ f'(x)=\frac{d}{dx}\left(x^7\right)e^{-\frac{x^2}{4}}+x^7\frac{d}{dx}\left(e^{-\frac{x^2}{4}}\right). \] Differentiate each part: 1. \(\frac{d}{dx}\left(x^7\right)=7x^6.\) 2. To differentiate \( e^{-\frac{x^2}{4}} \), use the chain rule. Let \( u(x)=-\frac{x^2}{4} \); then \[ \frac{du}{dx}=-\frac{x}{2} \quad\text{and}\quad \frac{d}{dx}\left(e^{u(x)}\right)=e^{u(x)}\frac{du}{dx}=-\frac{x}{2}e^{-\frac{x^2}{4}}. \] So, the derivative is \[ f'(x)=7x^6e^{-\frac{x^2}{4}}+x^7\left(-\frac{x}{2}e^{-\frac{x^2}{4}}\right). \] This simplifies to \[ f'(x)=e^{-\frac{x^2}{4}}\left(7x^6-\frac{x^8}{2}\right). \] ### Step 2. Factor the Derivative Factor \( x^6 \) from the expression: \[ f'(x)=e^{-\frac{x^2}{4}}x^6\left(7-\frac{x^2}{2}\right). \] Since \( e^{-\frac{x^2}{4}} \) is never zero, we set the remaining factors equal to zero: \[ x^6\left(7-\frac{x^2}{2}\right)=0. \] ### Step 3. Solve for Critical Points Set each factor equal to zero: 1. \( x^6=0 \) gives \[ x=0. \] 2. \( 7-\frac{x^2}{2}=0 \) gives \[ \frac{x^2}{2}=7, \] \[ x^2=14, \] \[ x=\sqrt{14} \quad \text{or} \quad x=-\sqrt{14}. \] Thus, the critical points are \[ x=-\sqrt{14}, \quad x=0, \quad x=\sqrt{14}. \] ### Step 4. Analyze the Nature of the Critical Points (Optional) - For \( x=-\sqrt{14} \): To the left of \( x=-\sqrt{14} \), \( 7-\frac{x^2}{2} \) is negative (since \( x^2>14 \)), making \( f'(x) \) negative. For \( x \) just to the right, \( 7-\frac{x^2}{2} \) is positive. Thus, \( f(x) \) changes from decreasing to increasing, indicating a **local minimum**. - For \( x=\sqrt{14} \): To the left, \( f'(x) \) is positive, and to the right, it is negative. Therefore, \( f(x) \) changes from increasing to decreasing, indicating a **local maximum**. - For \( x=0 \): The derivative \( f'(x) \) is zero, but around \( 0 \) (both to the left and right) \( 7-\frac{x^2}{2}>0 \), so \( f'(x) \) remains positive on both sides. Hence, \( x=0 \) is a **stationary point with a horizontal tangent** but not a local extremum. ### Final Answer The critical points of the function \[ f(x)=x^7e^{-\frac{x^2}{4}} \] are: - \( x=-\sqrt{14} \) (local minimum), - \( x=0 \) (a saddle point or point of inflection with a horizontal tangent), - \( x=\sqrt{14} \) (local maximum). These are the points to mark on the graph.