Question

find the critical numbers, the intervals on which f(x) is increasing, the intervals on which f(x) is decreasing, and the local extrema. do not graph.
f(x)=x + \frac{49}{x}
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the critical number(s) is(are)
(type integers or simplified fractions. use a comma to separate answers as needed.)
b. the function has no critical numbers.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function is increasing on
(type your answer in interval notation. type integers or simplified fractions. use a comma to separate answers as needed.)
b. the function is never increasing.
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the function is decreasing on

Explanation:

Step1: Find the derivative

First, rewrite \(f(x)=x + 49x^{-1}\). Then, using the power - rule \((x^n)^\prime=nx^{n - 1}\), we get \(f^\prime(x)=1-49x^{-2}=1-\frac{49}{x^{2}}=\frac{x^{2}-49}{x^{2}}=\frac{(x - 7)(x + 7)}{x^{2}}\).

Step2: Find critical numbers

Set \(f^\prime(x)=0\), then \(\frac{(x - 7)(x + 7)}{x^{2}}=0\). The numerator \((x - 7)(x + 7)=0\) gives \(x=7\) or \(x=-7\). Also, \(f(x)\) is undefined at \(x = 0\), but critical numbers are in the domain of the function. So the critical numbers are \(x=-7,7\).

Step3: Determine intervals of increase and decrease

We consider the intervals \((-\infty,-7)\), \((-7,0)\), \((0,7)\) and \((7,\infty)\).

• For \(x\in(-\infty,-7)\), let's test \(x=-8\). Then \(f^\prime(-8)=\frac{(-8 - 7)(-8 + 7)}{(-8)^{2}}=\frac{(-15)\times(-1)}{64}=\frac{15}{64}>0\), so \(f(x)\) is increasing on \((-\infty,-7)\).
• For \(x\in(-7,0)\), let's test \(x=-1\). Then \(f^\prime(-1)=\frac{(-1 - 7)(-1 + 7)}{(-1)^{2}}=\frac{(-8)\times6}{1}=-48<0\), so \(f(x)\) is decreasing on \((-7,0)\).
• For \(x\in(0,7)\), let's test \(x = 1\). Then \(f^\prime(1)=\frac{(1 - 7)(1 + 7)}{1^{2}}=\frac{(-6)\times8}{1}=-48<0\), so \(f(x)\) is decreasing on \((0,7)\).
• For \(x\in(7,\infty)\), let's test \(x = 8\). Then \(f^\prime(8)=\frac{(8 - 7)(8 + 7)}{8^{2}}=\frac{1\times15}{64}=\frac{15}{64}>0\), so \(f(x)\) is increasing on \((7,\infty)\).

Step4: Find local extrema

Since \(f(x)\) changes from increasing to decreasing at \(x=-7\), \(f(-7)=-7+\frac{49}{-7}=-7 - 7=-14\) is a local maximum.
Since \(f(x)\) changes from decreasing to increasing at \(x = 7\), \(f(7)=7+\frac{49}{7}=7 + 7=14\) is a local minimum.

Answer:

• Critical numbers: A. The critical number(s) is(are) \(-7,7\).
• Intervals of increase: A. The function is increasing on \((-\infty,-7),(7,\infty)\).
• Intervals of decrease: A. The function is decreasing on \((-7,0),(0,7)\).
• Local extrema: Local maximum is \(f(-7)=-14\) and local minimum is \(f(7)=14\).