Question

consider the function ( f(x)=4x + 7x^{-1} ). for this function there are four important intervals: ( (-infty,a ), ( a,b) ), ( (b,c ), and ( c,infty) ) where ( a ), and ( c ) are the critical numbers and the function is not defined at ( b ). find ( a ) and ( b ) and ( c ) for each of the following open intervals, tell whether ( f(x) ) is increasing or decreasing. ( (-infty,a) ): select an answer ( (a,b) ): select an answer ( (b,c) ): select an answer ( (c,infty) ) select an answer note that this function has no inflection points, but we can still consider its concavity. for each of the following intervals, tell whether ( f(x) ) is concave up or concave down. ( (-infty,b) ): select an answer ( (b,infty) ): select an answer question help: video message instructor

Explanation:

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

Given \(f(x)=4x + 7x^{-1}\), using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=4-7x^{-2}=4-\frac{7}{x^{2}}=\frac{4x^{2}-7}{x^{2}}\).

Step2: Find the critical numbers

Set \(f^\prime(x) = 0\), so \(\frac{4x^{2}-7}{x^{2}}=0\). Since \(x^{2}
eq0\) (when finding critical - numbers from the derivative equation \(f^\prime(x) = 0\)), we solve \(4x^{2}-7 = 0\). Then \(x^{2}=\frac{7}{4}\), and \(x=\pm\frac{\sqrt{7}}{2}\). The function \(y = f(x)\) is not defined when \(x = 0\) (because of the \(x^{-1}=\frac{1}{x}\) term).

Step3: Determine the sign of \(f^\prime(x)\) in each interval

• For the interval \((-\infty,-\frac{\sqrt{7}}{2})\), let \(x=-2\). Then \(f^\prime(-2)=4-\frac{7}{4}=\frac{16 - 7}{4}=\frac{9}{4}>0\).
• For the interval \((-\frac{\sqrt{7}}{2},0)\), let \(x =-\frac{1}{2}\). Then \(f^\prime(-\frac{1}{2})=4 - 28=-24<0\).
• For the interval \((0,\frac{\sqrt{7}}{2})\), let \(x=\frac{1}{2}\). Then \(f^\prime(\frac{1}{2})=4 - 28=-24<0\).
• For the interval \((\frac{\sqrt{7}}{2},\infty)\), let \(x = 2\). Then \(f^\prime(2)=4-\frac{7}{4}=\frac{16 - 7}{4}=\frac{9}{4}>0\).
• Now find the second - derivative \(f^{\prime\prime}(x)\). Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) with \(u = 4x^{2}-7\) (\(u^\prime = 8x\)) and \(v=x^{2}\) (\(v^\prime = 2x\)), we have \(f^{\prime\prime}(x)=\frac{8x\cdot x^{2}-(4x^{2}-7)\cdot2x}{x^{4}}=\frac{8x^{3}-8x^{3}+14x}{x^{4}}=\frac{14}{x^{3}}\).
• For the interval \((-\infty,0)\), let \(x=-1\). Then \(f^{\prime\prime}(-1)=-14<0\) (concave down).
• For the interval \((0,\infty)\), let \(x = 1\). Then \(f^{\prime\prime}(1)=14>0\) (concave up).

Answer:

\(A=-\frac{\sqrt{7}}{2}\), \(B = 0\), \(C=\frac{\sqrt{7}}{2}\)
\((-\infty,A)\): increasing
\((A,B)\): decreasing
\((B,C)\): decreasing
\((C,\infty)\): increasing
\((-\infty,B)\): concave down
\((B,\infty)\): concave up