Question

suppose that $f(x)=(1 + 7x)e^{6x}$. a. find all critical values of $f$. (if there are multiple values, enter them separated by commas. if there are no critical points, enter dne.) b. use interval notation to indicate where $f$ is increasing. (if there are multiple intervals, enter them separated by commas.) c. use interval notation to indicate where $f$ is decreasing. (if there are multiple intervals, enter them separated by commas.) d. list the $x$-coordinates of all local maxima of $f$. (if there are no local maxima, enter dne.) e. list the $x$-coordinates of all local minima of $f$. (if there are no local minima, enter dne.)

Explanation:

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

First, use the product - rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = 1 + 7x\) and \(v=e^{6x}\). \(u^\prime=7\) and \(v^\prime = 6e^{6x}\). Then \(f^\prime(x)=7e^{6x}+6(1 + 7x)e^{6x}=e^{6x}(7 + 6+42x)=e^{6x}(42x + 13)\).

Step2: Find critical points

Set \(f^\prime(x)=0\). Since \(e^{6x}\gt0\) for all real \(x\), we solve \(42x+13 = 0\). So \(x=-\frac{13}{42}\).

Step3: Determine intervals of increase and decrease

Choose a test - point in the interval \((-\infty,-\frac{13}{42})\), say \(x=-1\). Then \(f^\prime(-1)=e^{-6}(42\times(-1)+13)=e^{-6}(-29)\lt0\), so \(f(x)\) is decreasing on \((-\infty,-\frac{13}{42})\). Choose a test - point in the interval \((-\frac{13}{42},\infty)\), say \(x = 0\). Then \(f^\prime(0)=e^{0}(42\times0 + 13)=13\gt0\), so \(f(x)\) is increasing on \((-\frac{13}{42},\infty)\).

Step4: Find local extrema

Since \(f(x)\) changes from decreasing to increasing at \(x = -\frac{13}{42}\), there is a local minimum at \(x=-\frac{13}{42}\). There are no local maxima.

Answer:

a. \(-\frac{13}{42}\)
b. \((-\frac{13}{42},\infty)\)
c. \((-\infty,-\frac{13}{42})\)
d. DNE
e. \(-\frac{13}{42}\)