question 3 · 1 point\nfind the absolute maximum and absolute minimum of the function ( f(x)=\frac{x}{x^{2}+9}…

question 3 · 1 point\nfind the absolute maximum and absolute minimum of the function ( f(x)=\frac{x}{x^{2}+9} ) over the interval (-5,5).\nenter an exact answer. if there is more than one value of ( x ) in the interval at which the maximum or minimum occurs, you should use a comma to separate them.\nprovide your answer below:\n- absolute maximum of ( square ) at ( x=square )\n- absolute minimum of ( square ) at ( x=square )

question 3 · 1 point\nfind the absolute maximum and absolute minimum of the function ( f(x)=\frac{x}{x^{2}+9} ) over the interval (-5,5).\nenter an exact answer. if there is more than one value of ( x ) in the interval at which the maximum or minimum occurs, you should use a comma to separate them.\nprovide your answer below:\n- absolute maximum of ( square ) at ( x=square )\n- absolute minimum of ( square ) at ( x=square )

Answer

Explanation:

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

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = x), (u^\prime=1), (v=x^{2}+9), (v^\prime = 2x). [ \begin{align*} f^\prime(x)&=\frac{(x^{2}+9)\times1 - x\times(2x)}{(x^{2}+9)^{2}}\ &=\frac{x^{2}+9 - 2x^{2}}{(x^{2}+9)^{2}}\ &=\frac{9 - x^{2}}{(x^{2}+9)^{2}} \end{align*} ]

Step2: Find the critical points

Set (f^\prime(x)=0), so (\frac{9 - x^{2}}{(x^{2}+9)^{2}} = 0). Since ((x^{2}+9)^{2}>0) for all (x), we solve (9 - x^{2}=0), which gives (x=\pm3). Both (x = 3) and (x=- 3) are in the interval ([-5,5]).

Step3: Evaluate the function at critical points and endpoints

  • When (x=-5): (f(-5)=\frac{-5}{(-5)^{2}+9}=\frac{-5}{25 + 9}=-\frac{5}{34})
  • When (x=-3): (f(-3)=\frac{-3}{(-3)^{2}+9}=\frac{-3}{9 + 9}=-\frac{1}{6})
  • When (x = 3): (f(3)=\frac{3}{3^{2}+9}=\frac{3}{9+9}=\frac{1}{6})
  • When (x = 5): (f(5)=\frac{5}{5^{2}+9}=\frac{5}{25 + 9}=\frac{5}{34})

Answer:

  • Absolute maximum of (\frac{1}{6}) at (x = 3)
  • Absolute minimum of (-\frac{1}{6}) at (x=-3)