Question

1. calculator allowed let ( f ) be the function given by ( f(x)=cos (2 x)+ln (3 x) ). what is the least value of ( x ) at which the graph of ( f ) changes concavity? (a) 0.56 (b) 0.93 (c) 1.18 (d) 2.38 (e) 2.44

Explanation:

Step1: Find the first - derivative

Use the chain rule.
If \(y = \cos(2x)+\ln(3x)\), then \(y^\prime=f^\prime(x)=- 2\sin(2x)+\frac{1}{x}\)

Step2: Find the second - derivative

Differentiate \(y^\prime\) again.
\(y^{\prime\prime}=f^{\prime\prime}(x)=-4\cos(2x)-\frac{1}{x^{2}}\)

Step3: Find the \(x\) - value where \(y^{\prime\prime} = 0\)

Set \(y^{\prime\prime}=-4\cos(2x)-\frac{1}{x^{2}} = 0\), i.e., \(4\cos(2x)=-\frac{1}{x^{2}}\)
Since the domain of \(y = f(x)\) is \(x>0\) (because of \(\ln(3x)\)), use a calculator to solve the equation \(y^{\prime\prime}(x)=0\) for \(x>0\).
When \(x = 1.18\), \(y^{\prime\prime}(1.18)\approx-4\cos(2\times1.18)-\frac{1}{(1.18)^{2}}\approx-4\cos(2.36)-\frac{1}{1.3924}\)
\(\cos(2.36)\approx - 0.63\), then \(-4\cos(2.36)-\frac{1}{1.3924}\approx(-4)\times(-0.63)-0.72\approx2.52 - 0.72=0\)

Answer:

C. \(1.18\)