![f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1](https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%5E4%20%2B%203x%5E3%20-%202x%5E2%20-%206x%20-%201)
Lets check with every option
(a) [-4,-3]
We plug in -4 for x and -3 for x
![f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55](https://tex.z-dn.net/?f=f%28-4%29%20%3D%20%28-4%29%5E4%20%2B%203%28-4%29%5E3%20-%202%28-4%29%5E2%20-%206%28-4%29%20-%201%3D%2055)
![f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1](https://tex.z-dn.net/?f=f%28-3%29%20%3D%20%28-3%29%5E4%20%2B%203%28-3%29%5E3%20-%202%28-3%29%5E2%20-%206%28-3%29%20-%201%3D%20-1)
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(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1](https://tex.z-dn.net/?f=f%28-3%29%20%3D%20%28-3%29%5E4%20%2B%203%28-3%29%5E3%20-%202%28-3%29%5E2%20-%206%28-3%29%20-%201%3D%20-1)
![f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5](https://tex.z-dn.net/?f=f%28-2%29%20%3D%20%28-2%29%5E4%20%2B%203%28-2%29%5E3%20-%202%28-2%29%5E2%20-%206%28-2%29%20-%201%3D%20-5)
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) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5](https://tex.z-dn.net/?f=f%28-2%29%20%3D%20%28-2%29%5E4%20%2B%203%28-2%29%5E3%20-%202%28-2%29%5E2%20-%206%28-2%29%20-%201%3D%20-5)
![f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1](https://tex.z-dn.net/?f=f%28-1%29%20%3D%20%28-1%29%5E4%20%2B%203%28-1%29%5E3%20-%202%28-1%29%5E2%20-%206%28-1%29%20-%201%3D%201)
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) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1](https://tex.z-dn.net/?f=f%28-1%29%20%3D%20%28-1%29%5E4%20%2B%203%28-1%29%5E3%20-%202%28-1%29%5E2%20-%206%28-1%29%20-%201%3D%201)
![f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1](https://tex.z-dn.net/?f=f%280%29%20%3D%20%280%29%5E4%20%2B%203%280%29%5E3%20-%202%280%29%5E2%20-%206%280%29%20-%201%3D%20-1)
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) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1](https://tex.z-dn.net/?f=f%280%29%20%3D%20%280%29%5E4%20%2B%203%280%29%5E3%20-%202%280%29%5E2%20-%206%280%29%20-%201%3D%20-1)
![f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5](https://tex.z-dn.net/?f=f%281%29%20%3D%20%281%29%5E4%20%2B%203%281%29%5E3%20-%202%281%29%5E2%20-%206%281%29%20-%201%3D%20-5)
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(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5](https://tex.z-dn.net/?f=f%281%29%20%3D%20%281%29%5E4%20%2B%203%281%29%5E3%20-%202%281%29%5E2%20-%206%281%29%20-%201%3D%20-5)
![f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19](https://tex.z-dn.net/?f=f%282%29%20%3D%20%282%29%5E4%20%2B%203%282%29%5E3%20-%202%282%29%5E2%20-%206%282%29%20-%201%3D%2019)
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]