find the absolute minimum and absolute maximum values of ( f ) on the given interval.\n\n( f(x)=x-ln (5 x)…

find the absolute minimum and absolute maximum values of ( f ) on the given interval.\n\n( f(x)=x-ln (5 x), quadleft\frac{1}{2}, 2\right )\n\nabsolute minimum value\n\nabsolute maximum value\n\nresources\n\nread it

find the absolute minimum and absolute maximum values of ( f ) on the given interval.\n\n( f(x)=x-ln (5 x), quadleft\frac{1}{2}, 2\right )\n\nabsolute minimum value\n\nabsolute maximum value\n\nresources\n\nread it

Answer

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)))