Question

suppose that $f(x)=(8 - 2x)e^{x}$. note: several parts of this problem require answers entered in interval notation. note, with interval notation, you can enter the empty set as {}. (a) list all the critical values of $f(x)$. note: if there are no critical values, enter none. (b) use interval notation to indicate where $f(x)$ is increasing. increasing: (c) use interval notation to indicate where $f(x)$ is decreasing. decreasing: (d) list the $x$ values of all local maxima of $f(x)$. if there are no local maxima, enter none. $x$ values of local maxima = (e) list the $x$ values of all local minima of $f(x)$. if there are no local minima, enter none. $x$ values of local minima = (f) use interval notation to indicate where $f(x)$ is concave up. concave up: (g) use interval notation to indicate where $f(x)$ is concave down. concave down: (h) list the $x$ values of all the inflection points of $f$. if there are no inflection points, enter none.

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 8 - 2x\) and \(v=e^{x}\). \(u^\prime=- 2\) and \(v^\prime = e^{x}\). So \(f^\prime(x)=-2e^{x}+(8 - 2x)e^{x}=(6 - 2x)e^{x}\).

Step2: Find the critical values

Set \(f^\prime(x)=0\). Since \(e^{x}\gt0\) for all \(x\), we solve \(6 - 2x = 0\), which gives \(x = 3\).

Answer:

A. \(x = 3\)

Step3: Determine intervals of increase and decrease

Choose test - points. For \(x\lt3\), let \(x = 0\), then \(f^\prime(0)=(6-0)e^{0}=6\gt0\), so \(f(x)\) is increasing on \((-\infty,3)\). For \(x\gt3\), let \(x = 4\), then \(f^\prime(4)=(6 - 8)e^{4}=-2e^{4}\lt0\), so \(f(x)\) is decreasing on \((3,\infty)\).