Mathematics - Application of Derivative Question with Solution | TestHub

Given $f(x)=x{e}^{-a{x}^2}$. We found from the extremum condition at $x=1$ that $a=1/2$. So, $f(x)=x{e}^{-x^2/2}$. First derivative: $f^{\prime }(x)=e^{-x^2/2}(1-x^2)$. Second derivative: $f^{\prime \prime }(x)=-x{e}^{-x^2/2}(1-x^2)+e^{-x^2/2}(-2x)=e^{-x^2/2}(-x+x^3-2x)=e^{-x^2/2}(x^3-3x)=x(x^2-3)e^{-x^2/2}$.

Points of inflection occur where $f^{\prime \prime }(x)=0$ and $f^{\prime \prime }(x)$ changes sign. Setting $f^{\prime \prime }(x)=0$ gives $x(x^2-3)=0$, so $x=0,x=\sqrt{3},x=-\sqrt{3}$. The point of inflection with the largest positive x-coordinate is $x_0=\sqrt{3}$. At $x_0=\sqrt{3}$: The y-coordinate is $y_0=f(\sqrt{3})=\sqrt{3}{e}^{-(\sqrt{3}{)}^2/2}=\sqrt{3}{e}^{-3/2}$. The slope of the tangent $m=f^{\prime }(\sqrt{3})=e^{-(\sqrt{3}{)}^2/2}(1-(\sqrt{3}{)}^2)=e^{-3/2}(1-3)=-2e^{-3/2}$. The equation of the tangent line $L$ is $y-y_0=m(x-x_0)$:$y-\sqrt{3}{e}^{-3/2}=-2e^{-3/2}(x-\sqrt{3})$.

Since $L$ passes through the point $(k,0)$, we substitute these coordinates into the tangent equation:$0-\sqrt{3}{e}^{-3/2}=-2e^{-3/2}(k-\sqrt{3})$. Dividing by $e^{-3/2}$ (which is non-zero):$-\sqrt{3}=-2(k-\sqrt{3})$.$\sqrt{3}=2(k-\sqrt{3})$.$\frac{\sqrt{3}}{2}=k-\sqrt{3}$.$k=\sqrt{3}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$. Calculating the numerical value: $k=\frac{3\times 1.73205}{2}\approx \frac{5.19615}{2}\approx 2.598075$. Rounding to two decimal places, the value of $k$ is $2.60$.