Solve for x: (x^2+2x)+x+2=0
2 answers:
You might be interested in
Answer:
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.
