x | f(x)\n-2 | 0\n-1 | 45\n0 | 64\n1 | 45\n2 | 0\n3 | -35\n4 | 0\n5 | 189\n6 | 640\naccording to the table…

x | f(x)\n-2 | 0\n-1 | 45\n0 | 64\n1 | 45\n2 | 0\n3 | -35\n4 | 0\n5 | 189\n6 | 640\naccording to the table, which ordered pair is a local maximum of the function, f(x)?\n(0, 64)\n(3, -35)\n(5, 189)\n(2, 0)

x | f(x)\n-2 | 0\n-1 | 45\n0 | 64\n1 | 45\n2 | 0\n3 | -35\n4 | 0\n5 | 189\n6 | 640\naccording to the table, which ordered pair is a local maximum of the function, f(x)?\n(0, 64)\n(3, -35)\n(5, 189)\n(2, 0)

Answer

Explanation:

Step1: Recall the definition of local maximum

A local maximum of a function (y = f(x)) is a point ((a,f(a))) such that (f(a)\geq f(x)) for all (x) in some open interval containing (a).

Step2: Analyze the values around each (x) - value

  • For (x = 0): We look at the values of (f(x)) for (x=-1) and (x = 1). (f(-1)=45), (f(0) = 64), (f(1)=45). Since (45<64), (f(0)\geq f(x)) for (x=-1) and (x = 1) (in the interval ((-1,1))).
  • For (x = 3): (f(2)=0), (f(3)=- 35), (f(4)=0). Since (-35<0), it is not a local maximum.
  • For (x = 5): We need to check the values around (x = 5). But we only know (f(4) = 0) and (f(6)=640). Since (0<189<640), it is not a local maximum (because (f(6)>f(5))).
  • For (x = 2): (f(1)=45), (f(2)=0), (f(3)=-35). Since (0<45), it is not a local maximum.

Answer:

((0,64))