Question

the function ( f(x) = 5 + 9x + 144x^{-1} ) has one local minimum and one local maximum. this function has a local maximum at ( x = ) with value and a local minimum at ( x = ) with value

Explanation:

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

Given \(f(x)=5 + 9x+144x^{-1}\), using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=9-144x^{-2}=9-\frac{144}{x^{2}}=\frac{9x^{2}-144}{x^{2}}=\frac{9(x^{2}-16)}{x^{2}}=\frac{9(x - 4)(x + 4)}{x^{2}}\), \(x
eq0\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(\frac{9(x - 4)(x + 4)}{x^{2}}=0\). Since \(x
eq0\), we solve \(x-4 = 0\) or \(x + 4=0\). The critical points are \(x=-4\) and \(x = 4\).

Step3: Use the second - derivative test

First, find the second - derivative. \(f^\prime(x)=9-144x^{-2}\), then \(f^{\prime\prime}(x)=288x^{-3}=\frac{288}{x^{3}}\).

• When \(x=-4\), \(f^{\prime\prime}(-4)=\frac{288}{(-4)^{3}}=\frac{288}{-64}<0\). So \(x=-4\) is a local maximum.

Substitute \(x = - 4\) into \(f(x)\): \(f(-4)=5+9\times(-4)+144\times(-4)^{-1}=5-36 - 36=-67\).

• When \(x = 4\), \(f^{\prime\prime}(4)=\frac{288}{4^{3}}=\frac{288}{64}>0\). So \(x = 4\) is a local minimum.

Substitute \(x = 4\) into \(f(x)\): \(f(4)=5+9\times4+144\times4^{-1}=5 + 36+36=77\).

Answer:

The function has a local maximum at \(x=-4\) with value \(-67\) and a local minimum at \(x = 4\) with value \(77\).