AP DSC School Assistant Math Model Test 1

Concept:We are given a function of the form $f(x) = \left(\frac{3}{x^{2}}\right)^{x}$ for $x>0$. To find its local maximum value, we use logarithmic differentiation. Taking natural log simplifies the exponent, then we differentiate, set the derivative to zero, solve for $x$, and evaluate the function at that critical point.Explanation:Step 1: Write $f(x) = \left(\frac{3}{x^{2}}\right)^{x}$. Take natural log on both sides:$\ln f(x) = x \ln\left(\frac{3}{x^{2}}\right) = x(\ln 3 - 2\ln x)$.Step 2: Differentiate both sides with respect to $x$:$\frac{f'(x)}{f(x)} = (\ln 3 - 2\ln x) + x\left(-\frac{2}{x}\right) = \ln 3 - 2\ln x - 2$So $f'(x) = f(x)\bigl(\ln 3 - 2\ln x - 2\bigr)$.Step 3: For local maximum, set $f'(x)=0$. Since $f(x) > 0$, we have:$\ln 3 - 2\ln x - 2 = 0$$\Rightarrow 2\ln x = \ln 3 - 2 = \ln 3 - \ln e^{2} = \ln\!\left(\frac{3}{e^{2}}\right)$$\Rightarrow \ln x^{2} = \ln\!\left(\frac{3}{e^{2}}\right)$$\Rightarrow x^{2} = \frac{3}{e^{2}}$ $\Rightarrow x = \frac{\sqrt{3}}{e}$ (since $x>0$).Step 4: Substitute this $x$ back into $f(x)$:$f\!\left(\frac{\sqrt{3}}{e}\right) = \left(\frac{3}{\left(\frac{\sqrt{3}}{e}\right)^{2}}\right)^{\frac{\sqrt{3}}{e}} = \left(\frac{3}{\frac{3}{e^{2}}}\right)^{\frac{\sqrt{3}}{e}} = \left(e^{2}\right)^{\frac{\sqrt{3}}{e}} = e^{\frac{2\sqrt{3}}{e}}$.Thus the local maximum value is $e^{\frac{2\sqrt{3}}{e}}$, which matches option C.Answer:The local maximum value of $f(x)$ is $e^{\frac{2\sqrt{3}}{e}}$, i.e., option C.