Question

The local maximum value of the function \(f(x) = \left(\frac{2}{x}\right)^{x^2}, x>0\), is :

• A. \((2\sqrt{e})^{\frac{1}{e}}\)
• B. \((e)^{\frac{2}{e}}\)
• C. \((\sqrt[4]{e})^{\frac{4}{\sqrt{e}}}\)
• D. 1

Hint:

For functions of the form \(y = u(x)^{v(x)}\), always use logarithmic differentiation. It transforms the problem into differentiating a product, which is much simpler. The critical points are often found by setting the derivative of the logarithmic expression to zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Question
We need to find the local maximum value of a function of the form \((g(x))^{h(x)}\). The standard method for this is logarithmic differentiation.
Step 2: Key Formula or Approach
Let \(y = f(x)\). Take the natural logarithm of both sides to get \(\ln y = h(x) \ln(g(x))\). Differentiate implicitly to find \(y'\). Set \(y' = 0\) to find critical points. Use the first or second derivative test to confirm it's a maximum.
Step 3: Detailed Explanation
Logarithmic Differentiation:
Let \(y = \left(\frac{2}{x}\right)^{x^2}\).
Taking the natural log on both sides: \[ \ln y = x^2 \ln\left(\frac{2}{x}\right) = x^2(\ln 2 - \ln x) \] Differentiating with respect to \(x\): \[ \frac{1}{y} \frac{dy}{dx} = (2x)(\ln 2 - \ln x) + x^2\left(-\frac{1}{x}\right) \] \[ \frac{1}{y} \frac{dy}{dx} = 2x(\ln 2 - \ln x) - x = x(2(\ln 2 - \ln x) - 1) = x\left(2\ln\left(\frac{2}{x}\right) - 1\right) \] \[ \frac{dy}{dx} = y \cdot x \left(2\ln\left(\frac{2}{x}\right) - 1\right) = \left(\frac{2}{x}\right)^{x^2} \cdot x \left(2\ln\left(\frac{2}{x}\right) - 1\right) \] Find Critical Points:
Set \(\frac{dy}{dx} = 0\). Since \(x>0\) and \(\left(\frac{2}{x}\right)^{x^2}>0\), we must have: \[ 2\ln\left(\frac{2}{x}\right) - 1 = 0 \] \[ \ln\left(\frac{2}{x}\right) = \frac{1}{2} \] \[ \frac{2}{x} = e^{1/2} = \sqrt{e} \] \[ x = \frac{2}{\sqrt{e}} \] Verify Maximum:
Let \(g(x) = 2\ln\left(\frac{2}{x}\right) - 1\). The sign of \(f'(x)\) is determined by the sign of \(g(x)\). \(g'(x) = 2 \cdot \frac{x}{2} \cdot \left(-\frac{2}{x^2}\right) = -\frac{2}{x}\). Since \(x>0\), \(g'(x)<0\), which means \(g(x)\) is a decreasing function. It passes from positive to negative at the critical point \(x = 2/\sqrt{e}\). Therefore, \(f'(x)\) also changes from positive to negative, confirming a local maximum.
Calculate Maximum Value:
Substitute \(x = \frac{2}{\sqrt{e}}\) into \(f(x)\). \[ f\left(\frac{2}{\sqrt{e}}\right) = \left(\frac{2}{2/\sqrt{e}}\right)^{(2/\sqrt{e})^2} = (\sqrt{e})^{4/e} = (e^{1/2})^{4/e} = e^{\frac{1}{2} \cdot \frac{4}{e}} = e^{2/e} \] Step 4: Final Answer
The local maximum value is \(e^{2/e}\).