Question

answer the questions below about the function whose derivative is ( f^{prime}(x)=1-\frac{49}{x^{2}}, x
eq 0 ).
a. what are the critical points of ( f )?
b. on what open intervals is ( f ) increasing or decreasing?
c. at what points, if any, does ( f ) assume local maximum and minimum values?

Explanation:

Step1: Find critical points

Set \(f^{\prime}(x)=0\), so \(1-\frac{49}{x^{2}} = 0\).

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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 \(x = 7\).