answer the questions below about the function whose derivative is ( f^{prime}(x)=1-\frac{49}{x^{2}}, x \neq…

answer the questions below about the function whose derivative is ( f^{prime}(x)=1-\frac{49}{x^{2}}, x \neq 0 ).\na. what are the critical points of ( f )?\nb. on what open intervals is ( f ) increasing or decreasing?\nc. at what points, if any, does ( f ) assume local maximum and minimum values?
Answer
Explanation:
Step1: Find critical points
Set (f^{\prime}(x)=0), so (1-\frac{49}{x^{2}} = 0). [ \begin{align*} 1-\frac{49}{x^{2}}&=0\ \frac{x^{2}-49}{x^{2}}&=0\ x^{2}-49&=0\ x^{2}&=49\ x&=\pm7 \end{align*} ]
Step2: Determine intervals of increase/decrease
Choose test points:
- For (x < - 7) (e.g., (x=-8)), (f^{\prime}(-8)=1-\frac{49}{64}=\frac{15}{64}>0), so (f(x)) is increasing on ((-\infty,-7)).
- For (-7 < x < 0) (e.g., (x = - 1)), (f^{\prime}(-1)=1 - 49=-48<0).
- For (0 < x < 7) (e.g., (x = 1)), (f^{\prime}(1)=1 - 49=-48<0).
- For (x>7) (e.g., (x = 8)), (f^{\prime}(8)=1-\frac{49}{64}=\frac{15}{64}>0), so (f(x)) is increasing on ((7,\infty)).
Step3: Find local maxima/minima
Since (f^{\prime}(x)) changes from positive to negative at (x=-7), (f(-7)) is a local maximum. Since (f^{\prime}(x)) changes from negative to positive at (x = 7), (f(7)) is a local minimum.
Answer:
a. The critical points are (x=-7) and (x = 7). b. (f(x)) is increasing on ((-\infty,-7)\cup(7,\infty)) and decreasing on ((-7,0)\cup(0,7)). c. Local maximum at (x=-7) and local minimum at (x = 7).