90000 1000 + 7000 + 7 =
2 answers:
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Lets check with every option
(a) [-4,-3]
We plug in -4 for x and -3 for x
f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.
(b) [-3,-2]
We plug in -3 for x and -2 for x
f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.
(c) [-2,-1]
We plug in -2 for x and -1 for x
f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.
(d) [-1,0]
We plug in -1 for x and 0 for x
f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.
(e) [0,1]
We plug in 0 for x and 1 for x
f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.
(f) [1,2]
We plug in 1 for x and 2 for x
f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.
so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]