# bounds on the real zerosFind the bounds on the real zeros of the following function. Please show all of your work.

Asked on by deblove13

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should use the first derivative to check what the zeroes of the function are.

Differentiating with respect to x yields:

`f'(x) = 4x^3 + 6x^2 - 2x`

You need to solve the equation `4x^3 + 6x^2 - 2x = 0` , hence factoring out 2x yields:

`2x(2x^2 + 3x - 1) = 0`

`` `2x = 0 =gt x_1 = 0`

You need to find the zeroes of `2x^2 + 3x - 1 = 0` , hence you may use quadratic formula such that:

`x_(2,3) = (-3 +- sqrt(9 + 8))/4 =gt x_2 = (-3+sqrt17)/4 ~~0.28`

`x_3 = (-3-sqrt17)/4 ~~ - 1.78`

You need to evaluate the value of the function at x = -1.78 such that:

`f(-1.78) = (-1.78)^4+2(-1.78)^3-(-1.78)^2-1`

`f(-1.78) = 10.038 - 11.279 - 3.168 - 1 = -5.409`

You need to evaluate the value of the function at x = 0 such that:

f(0) = -1

You need to remember that between two consecutive roots of derivative there is a root of the function if the sign of value of the function at `x_1` (one root of derivative) is distinct compared to the sign value of the function at `x_2` (next root of derivative).

Since the values of f(-1.78) and f(0) have like signs, then there is no real root between x=-1.78 and x=0.

You need to evaluate the value of the function at x = 0.28 such that:

`f(0.28) = (0.28)^4+2(0.28)^3-(0.28)^2-1`

`f(0.28) = 0.006 + 0.125 - 0.078 - 1 = -0.947`

Since the values of f(0) and f(0.28) have like signs, then there is no real root between x=0 and x=0.28.

Notice that the graph intercepts x axis at a value that is close to 1, hence larger than 0.28, then the function has a real root in interval (0.28,1).

Notice that the graph intercepts x axis at a value comprised in (-3,-2), hence the function has a real root in interval (-3,-2).

Hence, the function has only two real roots, one in  (-3,-2) and the other in (0.28,1), meaning that the other two roots are complex.

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