Question

the function $f(x)=(6x - 1)e^{-4x}$ has one critical number. find the critical number.

Explanation:

Step1: Recall the product - rule for differentiation

If \(y = uv\), where \(u\) and \(v\) are functions of \(x\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = 6x - 1\) and \(v=e^{-4x}\). Then \(u^\prime=6\) and \(v^\prime=- 4e^{-4x}\).

Step2: Calculate the derivative of \(f(x)\)

Using the product - rule \(f^\prime(x)=u^\prime v+uv^\prime\), we have \(f^\prime(x)=6e^{-4x}+(6x - 1)(-4e^{-4x})\).
\[

$$\begin{align*} f^\prime(x)&=6e^{-4x}-24xe^{-4x}+4e^{-4x}\\ &=(6 + 4-24x)e^{-4x}\\ &=(10 - 24x)e^{-4x} \end{align*}$$

\]

Step3: Find the critical number

Critical numbers occur where \(f^\prime(x) = 0\). Since \(e^{-4x}
eq0\) for all real \(x\) (the exponential function \(y = e^t\) is never zero for any real \(t\), and here \(t=-4x\)), we set \(10 - 24x = 0\).
\[

$$\begin{align*} 10-24x&=0\\ 24x&=10\\ x&=\frac{10}{24}=\frac{5}{12} \end{align*}$$

\]

Answer:

\(x = \frac{5}{12}\)