An interval over which the function, f(x) = -2x³ - 3x +5 is guaranteed to have a zero is [0,2]
Further explanation
If equation ax³ + bx² + cx + d = 0 has roots x₁ , x₂ , and x₃ then
[tex]x_1 + x_2 + x_3 = - \frac{b}{a}[/tex]
[tex]x_1 x_2 + x_1 x_3 + x_2 x_3 = \frac{c}{a}[/tex]
[tex]x_1 x_2 x_3 = - \frac{d}{a}[/tex]
Let us now tackle the problem!
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Given:
[tex]f(x) = -2x^3 - 3x + 5[/tex]
Option A:
[tex]f(x) = -2x^3 - 3x + 5[/tex]
[tex]f(-3) = -2(-3)^3 - 3(-3) + 5 = 68[/tex]
[tex]f(-2) = -2(-2)^3 - 3(-2) + 5 = 27[/tex]
Because both of the value of f(-3) and f(-2) are positive , we cannot guarateed the value of the function will be zero at interval [-3,-2]
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Option B:
[tex]f(x) = -2x^3 - 3x + 5[/tex]
[tex]f(0) = -2(0)^3 - 3(0) + 5 = 5[/tex]
[tex]f(-2) = -2(-2)^3 - 3(-2) + 5 = 27[/tex]
Because both of the value of f(0) and f(-2) are positive , we cannot guarateed the value of the function is zero at interval [-3,-2]
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Option C:
[tex]f(x) = -2x^3 - 3x + 5[/tex]
[tex]f(0) = -2(0)^3 - 3(0) + 5 = 5[/tex]
[tex]f(2) = -2(2)^3 - 3(2) + 5 = -17[/tex]
Because the value of f(0) and f(2) have different sign , we can guarateed the value of the function will be zero at interval [0,2] , i.e. there is zero between 5 and -17 → -17 < 0 < 5
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Option D:
[tex]f(x) = -2x^3 - 3x + 5[/tex]
[tex]f(4) = -2(4)^3 - 3(4) + 5 = -135[/tex]
[tex]f(2) = -2(2)^3 - 3(2) + 5 = -17[/tex]
Because both of the value of f(4) and f(2) are negatve , we cannot guarateed the value of the function is zero at interval [2,4]
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Learn more
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Answer details
Grade: High School
Subject: Mathematics
Chapter: Polynomial
Keywords: Quadratic , Equation , Discriminant , Real , Number
