Question
Use the Intemediale Walue Theorem to show that the polynomial f(x) \( =2 x^{4}-7 x^{2}+1 \) has a real zero between -1 and 0. Select the conect choice below and fill in the answer bowes to complete your choice. A. Because f(x) is a polynomial with f( -1 )= \( \square \) \( <0 \) and \( f(0)= \) \( \square \) \( >0 \), the function has a real zero between -1 and 0. B. Because f(x) is a polynomial with fi( -1 )= \( \square \) \( >0 \) and \( 7(\mathrm{D})= \) \( \square \) \( >0 \), the function has a real zero between -1 ando. C. Because f(x) is a potynomial with \( f(-1)= \) \( \square \) \( <0 \) and \( f(0)= \) \( \square \) \( <0 \), the funcion has a real zero between -1 and 0. D. Because \( ( \) (a) is a polynomial with \( \mathrm{f}(-1)= \) \( \square \) \( >0 \) and \( F(0)= \) \( \square \) \( <0 \), the function has a real zero between - 1 and 0.
Ask by Weber Campos. in the United States
Feb 14,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
A. Because f(x) is a polynomial with f(–1) = –4 < 0 and f(0) = 1 > 0, the function has a real zero between –1 and 0.
Solution
We first calculate:
f(–1) = 2(–1)⁴ – 7(–1)² + 1 = 2(1) – 7(1) + 1 = 2 – 7 + 1 = –4,
f(0) = 2(0)⁴ – 7(0)² + 1 = 1.
Since f(–1) = –4 < 0 and f(0) = 1 > 0, the Intermediate Value Theorem guarantees at least one real zero between –1 and 0.
Thus, the correct choice is:
A. Because f(x) is a polynomial with f(–1) = –4 < 0 and f(0) = 1 > 0, the function has a real zero between –1 and 0.
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Let's evaluate the polynomial \( f(x) = 2x^4 - 7x^2 + 1 \) at the points -1 and 0. First, for \( f(-1) \): \[ f(-1) = 2(-1)^4 - 7(-1)^2 + 1 = 2(1) - 7(1) + 1 = 2 - 7 + 1 = -4. \] Next, for \( f(0) \): \[ f(0) = 2(0)^4 - 7(0)^2 + 1 = 0 - 0 + 1 = 1. \] With these evaluations: - \( f(-1) = -4 < 0 \) - \( f(0) = 1 > 0 \) Thus, the correct choice is: A. Because f(x) is a polynomial with \( f(-1)= -4 < 0 \) and \( f(0)= 1 > 0 \), the function has a real zero between -1 and 0.
