Question

find the absolute minimum and absolute maximum values of ( f ) on the given interval.

( f(x)=x-ln (5 x), quadleft\frac{1}{2}, 2
ight )

absolute minimum value

absolute maximum value

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Explanation:

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

The derivative of \( f(x)=x - \ln(5x)\) is \( f^\prime(x)=1-\frac{1}{x}\) (using the rules \((x^n)^\prime = nx^{n - 1}\) and \((\ln u)^\prime=\frac{u^\prime}{u}\), here \(u = 5x\) and \(u^\prime=5\)).

Step2: Find the critical points

Set \( f^\prime(x)=0\), so \(1-\frac{1}{x}=0\). Solving for \(x\) gives \(x = 1\). Also, check the endpoints \(x=\frac{1}{2}\) and \(x = 2\).

Step3: Evaluate \( f(x) \) at critical points and endpoints

• For \(x=\frac{1}{2}\): \(f(\frac{1}{2})=\frac{1}{2}-\ln(\frac{5}{2})\approx\frac{1}{2}-0.916=-0.416\)
• For \(x = 1\): \(f(1)=1-\ln(5)\approx1 - 1.609=-0.609\)
• For \(x = 2\): \(f(2)=2-\ln(10)\approx2-2.303=-0.303\)

Answer:

absolute minimum value: \(-0.609\) (approximate value of \(1-\ln(5)\))
absolute maximum value: \(-0.303\) (approximate value of \(2-\ln(10)\))