Carry out three steps of the bisection method for f(x)=3x−x4 as follows: (a) show that f(x) has a zero in [1,2]. (b) determine which subinterval, [1,1.5] or [1.5,2], contains a zero. (c) determine which interval, [1,1.25], [1.25,1.5], [1.5,1.75], or [1.75,2], contains a zero.

a) There's a zero between [1,2]

b) There's a zero between [1.5,2]

c) There's a zero between  [1.5,1.75].

Step-by-step explanation:

We have

A)We need to show that f(x) has a zero in the interval [1, 2]. We have to see if the function f is continuous with f(1) and f(2).

We can see that f(1) and f(2) have opposite signs. And f(1)>f(2) and the function is continuous, this means that exists a real number c between the interval [1,2] where f(c)=0.

B)We have to repeat the same steps of A)

For the subinterval [1,1.5]:

f(1) and f(1.5) have the same signs, this means there's no zero in the subinterval [1,1.5].

For the subinterval [1.5,2]:

f(1.5) and f(2) have opposite signs, this means there's a zero between the subinterval [1.5,2].

C)We have to repeat the same steps of A)

For the subinterval [1,1.25]:

f(1) and f(1.25) have the same signs, this means there's no zero in the subinterval [1,1.25].

For the subinterval [1.25,1.5]:

f(1.25) and f(1.5) have the same signs, this means there's no zero in the subinterval [1.25,1.5].

For the subinterval [1.5,1.75]:

f(1.5) and f(1.75) have opposite signs, this means there's a zero between the subinterval [1.5,1.75].

For the subinterval [1.75,2]:

f(1.75) and f(2) have the same signs, this means there isn't a zero between the subinterval [1.75,2].

The graph of the function shows that the answers are correct.

c) yes, because of asa or aas