contestada

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
x4 + x ? 7 = 0, (1, 2)
f(x) = x4 + x ? 7
is (FILL IN)
a) defined
b) continuous
c) negative
d) positive on the closed interval [1, 2],
f(1) = ?? FILL IN , and f(2) = ?? FILL IN
Since ?5 < FILL IN a)? b)? c)? d)0 < 11, there is a number c in (1, 2) such that
f(c) = FILL IN a)? b)? c)0 d)11 e)-5
by the Intermediate Value Theorem. Thus, there is a FILL IN a) limit b)root c) discontinuity of the equation
x4 + x ? 7 = 0
in the interval (1, 2).

Respuesta :

okpalawalter8

Answer:

The correct option is  d

[tex]f(1) = -5[/tex]

[tex]f(2) = 11[/tex]

The correct option is d

The correct option is  c

the correct option is  b

Step-by-step explanation:

The given equation is

     [tex]f(x) = x^4 + x -7 =0[/tex]

The give interval  is [tex](1,2)[/tex]

 

   Now differentiating the equation

        [tex]f'(x) = 4x^3 +7 > 0[/tex]

Therefore the equation is  positive  at the given interval

       Now at x= 1

  [tex]f(1) = (1)^4 + 1 -7 =-5[/tex]

       Now at x= 2

  [tex]f(2) = (2)^4 + 2 -7 =11[/tex]

Now at  the interval (1,2)

      [tex]f(1) < 0 < f(2)[/tex]

i.e

      [tex]-5 < 0 < 11[/tex]

this tell us that there is a value z within 1,2 and

   f(z) =  0

Which implies that there is a root within (1,2) according to the intermediate value theorem

     

Otras preguntas