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

find the absolute maximum and absolute minimum values of ( f ) on the given interval.\n\n( f(x)=ln left(x^{2}+5 x + 9\right), quad-3,1 )\n\nabsolute minimum value\n\nabsolute maximum value\n\nresources\n\nread it

find the absolute maximum and absolute minimum values of ( f ) on the given interval.\n\n( f(x)=ln left(x^{2}+5 x + 9\right), quad-3,1 )\n\nabsolute minimum value\n\nabsolute maximum value\n\nresources\n\nread it

Answer

Explanation:

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

Using the chain rule, if (y = \ln(u)) and (u=x^{2}+5x + 9), then (y^\prime=\frac{u^\prime}{u}). (u^\prime = 2x + 5), so (f^\prime(x)=\frac{2x + 5}{x^{2}+5x + 9})

Step2: Find the critical points

Set (f^\prime(x)=0), so (\frac{2x + 5}{x^{2}+5x + 9}=0). Since (x^{2}+5x + 9=(x+\frac{5}{2})^{2}+\frac{11}{4}>0) for all (x), then (2x+5 = 0), (x=-\frac{5}{2})

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

  • For (x=-3): (f(-3)=\ln((-3)^{2}+5\times(-3)+9)=\ln(9 - 15 + 9)=\ln(3))
  • For (x=-\frac{5}{2}): (f(-\frac{5}{2})=\ln((-\frac{5}{2})^{2}+5\times(-\frac{5}{2})+9)=\ln(\frac{25}{4}-\frac{25}{2}+9)=\ln(\frac{25 - 50 + 36}{4})=\ln(\frac{11}{4}))
  • For (x = 1): (f(1)=\ln(1^{2}+5\times1+9)=\ln(15))

Step4: Compare the values

(\ln(\frac{11}{4})\approx0.74), (\ln(3)\approx1.10), (\ln(15)\approx2.71)

Answer:

absolute minimum value: (\ln(\frac{11}{4})) absolute maximum value: (\ln(15))