Consider the following function. Use a graphing utility to conflm your answers for parts (a) through (c). (If an answer does not exist, enter DNE.] \[ f(x)=7 x+\frac{1}{x} \] (a) Find the critical numbers of \( f \). (Enter your answers as a comma-separated list.) \[ x=\square \] (b) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.) increasing \( \square \) decreasing \( \square \) (c) Apply the Frst Derivative Test to identify all relative extrema. relative maximum \( (x, y)=( \) \( \square \) \[ 1 \] relatlve minimum \( \quad(x, y)=( \) \( \square \) )

**(a)** To find the critical numbers, compute the derivative of \[ f(x)=7x+\frac{1}{x}. \] Differentiate term‐by‐term: \[ f'(x)=7-\frac{1}{x^2}. \] Critical numbers occur where \(f'(x)=0\) or is undefined (and \(x\) is in the domain of \(f\)). Here, \(f'(x)\) is undefined at \(x=0\), but since \(0\) is not in the domain of \(f(x)\), we ignore it. Set the derivative equal to zero: \[ 7-\frac{1}{x^2}=0\quad\Longrightarrow\quad \frac{1}{x^2}=7\quad\Longrightarrow\quad x^2=\frac{1}{7}. \] Thus, \[ x=\frac{1}{\sqrt{7}} \quad \text{or} \quad x=-\frac{1}{\sqrt{7}}. \] **(b)** The sign of \(f'(x)=7-\frac{1}{x^2}\) will determine where \(f\) is increasing or decreasing. The domain is split by the critical numbers and the discontinuity at \(x=0\) into the intervals: \[ (-\infty,-\frac{1}{\sqrt{7}}),\quad \left(-\frac{1}{\sqrt{7}},0\right),\quad \left(0,\frac{1}{\sqrt{7}}\right),\quad \left(\frac{1}{\sqrt{7}},\infty\right). \] 1. For \(x\) in \((-\infty,-\frac{1}{\sqrt{7}})\), choose \(x=-1\) (since \(-1<-\frac{1}{\sqrt{7}}\approx -0.377\)). Then, \[ f'(-1)=7-\frac{1}{1}=6>0. \] \(f\) is increasing on \((-\infty,-\frac{1}{\sqrt{7}})\). 2. For \(x\) in \(\left(-\frac{1}{\sqrt{7}},0\right)\), choose \(x=-0.1\). Then, \[ f'(-0.1)=7-\frac{1}{0.01}=7-100=-93<0. \] \(f\) is decreasing on \(\left(-\frac{1}{\sqrt{7}},0\right)\). 3. For \(x\) in \(\left(0,\frac{1}{\sqrt{7}}\right)\), choose \(x=0.1\). Then, \[ f'(0.1)=7-\frac{1}{0.01}=7-100=-93<0. \] \(f\) is decreasing on \(\left(0,\frac{1}{\sqrt{7}}\right)\). 4. For \(x\) in \(\left(\frac{1}{\sqrt{7}},\infty\right)\), choose \(x=1\). Then, \[ f'(1)=7-\frac{1}{1}=6>0. \] \(f\) is increasing on \(\left(\frac{1}{\sqrt{7}},\infty\right)\). Thus, in interval notation: - Increasing on: \[ (-\infty,-\frac{1}{\sqrt{7}})\cup \left(\frac{1}{\sqrt{7}},\infty\right) \] - Decreasing on: \[ \left(-\frac{1}{\sqrt{7}},0\right)\cup \left(0,\frac{1}{\sqrt{7}}\right) \] **(c)** Apply the First Derivative Test at the critical numbers. - At \(x=-\frac{1}{\sqrt{7}}\): \(f'(x)\) changes from positive (increasing) to negative (decreasing), indicating a relative maximum. Compute: \[ f\left(-\frac{1}{\sqrt{7}}\right)=7\left(-\frac{1}{\sqrt{7}}\right)+\frac{1}{-\frac{1}{\sqrt{7}}}=-\frac{7}{\sqrt{7}}-\sqrt{7}=-\sqrt{7}-\sqrt{7}=-2\sqrt{7}. \] Hence, the relative maximum is at \[ \left(-\frac{1}{\sqrt{7}},-2\sqrt{7}\right). \] - At \(x=\frac{1}{\sqrt{7}}\): \(f'(x)\) changes from negative (decreasing) to positive (increasing), indicating a relative minimum. Compute: \[ f\left(\frac{1}{\sqrt{7}}\right)=7\left(\frac{1}{\sqrt{7}}\right)+\frac{1}{\frac{1}{\sqrt{7}}}=\frac{7}{\sqrt{7}}+\sqrt{7}=\sqrt{7}+\sqrt{7}=2\sqrt{7}. \] Hence, the relative minimum is at \[ \left(\frac{1}{\sqrt{7}},2\sqrt{7}\right). \] **Final Answers** (a) \[ x=-\frac{1}{\sqrt{7}},\frac{1}{\sqrt{7}} \] (b) - Increasing: \[ (-\infty,-\frac{1}{\sqrt{7}})\cup \left(\frac{1}{\sqrt{7}},\infty\right) \] - Decreasing: \[ \left(-\frac{1}{\sqrt{7}},0\right)\cup \left(0,\frac{1}{\sqrt{7}}\right) \] (c) - Relative maximum at \(\left(-\frac{1}{\sqrt{7}},-2\sqrt{7}\right)\) - Relative minimum at \(\left(\frac{1}{\sqrt{7}},2\sqrt{7}\right)\)