Question

use the derivative to find and classify a critical number of $f(x) = \frac{x - 6}{e^x}$.

• function $f$ has select an answer at $x = \square$

Explanation:

Step1: Find the derivative of \( f(x) \)

Using the quotient rule: if \( f(x)=\frac{u(x)}{v(x)} \), then \( f^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v(x)^2} \). Here, \( u(x)=x - 6 \), \( u^\prime(x)=1 \); \( v(x)=e^x \), \( v^\prime(x)=e^x \). So \( f^\prime(x)=\frac{1\cdot e^x-(x - 6)e^x}{(e^x)^2}=\frac{e^x - (x - 6)e^x}{e^{2x}}=\frac{e^x(1-(x - 6))}{e^{2x}}=\frac{7 - x}{e^x} \).

Step2: Find critical numbers

Set \( f^\prime(x)=0 \) or \( f^\prime(x) \) undefined. \( e^x
eq0 \) for all \( x \), so solve \( 7 - x = 0 \), we get \( x = 7 \).

Step3: Classify the critical number

Use the first - derivative test. For \( x\lt7 \), say \( x = 0 \), \( f^\prime(0)=\frac{7-0}{e^0}=7\gt0 \), function is increasing. For \( x\gt7 \), say \( x = 8 \), \( f^\prime(8)=\frac{7 - 8}{e^8}=\frac{-1}{e^8}\lt0 \). So at \( x = 7 \), the function has a local maximum.

Answer:

Function \( f \) has a local maximum at \( x=\boldsymbol{7} \)