Option B: [tex]-1<x<0[/tex] is the interval in which [tex]f(x)[/tex] has a real zero
Explanation:
The given equation is [tex]f(x)=3 x^{3}-5 x^{2}+5 x+7[/tex]
We need to determine x at which the value of f(x) becomes zero.
Option A: [tex]-1<x<0 ; 0<x<1 ; 1<x<2 ; 2<x<3[/tex]
Let us substitute the values of x in the equation f(x), we get,
(i) Consider the 1st interval [tex]-1<x<0[/tex]
[tex]f(-1)=3 (-1)-5 (1)+5(-1)+7=-6[/tex]
[tex]f(0)=3 (0)-5 (0)+5(0)+7=7[/tex]
Since, there is a change of sign between the two interval, f(x) has a zero between the interval [tex]-1<x<0[/tex]
(ii) Consider the 2nd interval [tex]0<x<1[/tex]
[tex]f(0)=3 (0)-5 (0)+5(0)+7=7[/tex]
[tex]f(1)=3 (1)-5 (1)+5(1)+7=10[/tex]
Since, there is no change of sign between the two interval, f(x) does not have a zero between the interval [tex]0<x<1[/tex]
(iii) Consider the 3rd interval [tex]1<x<2[/tex]
[tex]f(1)=3 (1)-5 (1)+5(1)+7=10[/tex]
[tex]f(2)=3 (8)-5 (4)+5(2)+7=21[/tex]
Since, there is no change of sign between the two interval, f(x) does not have a zero between the interval [tex]1<x<2[/tex]
(iv) Consider the 4th interval [tex]2<x<3[/tex]
[tex]f(2)=3 (8)-5 (4)+5(2)+7=21[/tex]
[tex]f(3)=3 (27)-5 (9)+5(3)+7=58[/tex]
Since, there is no change of sign between the two interval, f(x) does not have a zero between the interval [tex]2<x<3[/tex]
From all the above 4 options, there is no change of sign between the intervals and hence, Option A is not the correct answer.
Option B: [tex]-1<x<0[/tex]
Let us substitute the values of x in the equation f(x), we get,
[tex]f(-1)=3 (-1)-5 (1)+5(-1)+7=-6[/tex]
[tex]f(0)=3 (0)-5 (0)+5(0)+7=7[/tex]
Since, there is a change of sign between the two interval, f(x) has a zero between the interval [tex]-1<x<0[/tex]
Hence, Option B is the correct answer.
Option C: [tex]1<x<2[/tex]
Let us substitute the values of x in the equation f(x), we get,
[tex]f(1)=3 (1)-5 (1)+5(1)+7=10[/tex]
[tex]f(2)=3 (8)-5 (4)+5(2)+7=21[/tex]
Since, there is no change of sign between the two interval, f(x) does not have a zero between the interval [tex]1<x<2[/tex]
Hence, Option C is not the correct answer.
Option D: [tex]-8<x<-7[/tex]
Let us substitute the values of x in the equation f(x), we get,
[tex]f(-8)=3 (-512)-5 (64)+5(-8)+7=-1889[/tex]
[tex]f(-7)=3 (-343)-5 (49)+5(-7)+7=-1302[/tex]
Since, there is no change of sign between the two interval, f(x) does not have a zero between the interval [tex]-8<x<-7[/tex]
Hence, Option D is not the correct answer.
