How do you find the relative extrema for f(x) = (x)3^-x?

Aug 3, 2017

${f}_{\max} = f \left(\cong 0.910\right) \approx 0.335$

Explanation:

$f \left(x\right) = x \cdot {3}^{- x}$

$f ' \left(x\right) = x \cdot \frac{d}{\mathrm{dx}} {3}^{- x} + {3}^{- x} \cdot 1$ [Product rule]

$\frac{d}{\mathrm{dx}} {3}^{- x} = - \ln 3 \cdot {3}^{- x}$

$\therefore f ' \left(x\right) = x \cdot \left(- \ln 3 \cdot {3}^{- x}\right) + {3}^{- x}$

$= - {3}^{- x} \left(x \ln 3 - 1\right)$

For critical points of $f \left(x\right) , f ' \left(x\right) = 0$

Since $- {3}^{- x} \ne 0 \forall x \in \mathbb{R}$

$x \ln 3 - 1 = 0$

$x = \frac{1}{\ln} 3 \approx 0.910$

$\therefore f \left(\approx 0.910\right)$ will be a critical value of $f \left(x\right)$

To determine whether the critical point is a local maximum or minimum.

$f ' ' \left(x\right) = - {3}^{- x} \cdot \ln 3 + \left(x \ln 3 - 1\right) \cdot \ln 3 \cdot {3}^{- x}$

$= - {3}^{- x} \ln 3 \left(1 - x \ln 3 + 1\right)$

$= - {3}^{- x} \ln 3 \left(2 - x \ln 3\right)$

Consider: $- {3}^{- x} \ln 3 < 0 \forall x \in \mathbb{R}$

And: $2 - \left(\cong 0.910\right) \cdot \ln 3 = \left(2 - \cong 1.00\right) > 0$

$\therefore f ' ' \left(\cong 0.910\right) < 0 \to f \left(\cong 0.910\right)$ is a local maximum

${f}_{\max} = f \left(\cong 0.910\right) \approx 0.335$

This can be seen on the graph of $f \left(x\right)$ below.

graph{ x*3^(-x) [-1.415, 5.515, -1.498, 1.966]}