On which interval does the Intermediate Value Theorem guarantee that the polynomial x^4 + 7x^2 − 9x −
1 has a root?
A. (-1/2,0)
B. (1/2,1)
C. (0,1/2)
D. (-1,-1/2)
To determine the interval on which the Intermediate Value Theorem guarantees that the polynomial x^4 + 7x^2 − 9x − 1 has a root, we can utilize the fact that the Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b] and takes on two different values, f(a) and f(b), then it must also take on every value in between.
In this case, the polynomial x^4 + 7x^2 − 9x − 1 is a continuous function, as polynomials are continuous for all values of x. Our goal is to find two values, a and b, such that f(a) and f(b) have opposite signs, indicating that the polynomial crosses the x-axis and has at least one root in between.
To find these two values, we can evaluate the polynomial at different points within each interval and observe the signs of the resulting values.
A. (-1/2,0):
f(-1/2) = (-1/2)^4 + 7(-1/2)^2 − 9(-1/2) − 1 = 1/16 + 7/4 + 9/2 − 1 = 17/16 > 0
f(0) = 0^4 + 7(0)^2 − 9(0) − 1 = -1 < 0
B. (1/2,1):
f(1/2) = (1/2)^4 + 7(1/2)^2 − 9(1/2) − 1 = 1/16 + 7/4 − 9/2 − 1 = -15/16 < 0
f(1) = 1^4 + 7(1)^2 − 9(1) − 1 = -2 < 0
C. (0,1/2):
f(0) = 0^4 + 7(0)^2 − 9(0) − 1 = -1 < 0
f(1/2) = (1/2)^4 + 7(1/2)^2 − 9(1/2) − 1 = 1/16 + 7/4 − 9/2 − 1 = -15/16 < 0
D. (-1,-1/2):
f(-1) = (-1)^4 + 7(-1)^2 − 9(-1) − 1 = -2 < 0
f(-1/2) = (-1/2)^4 + 7(-1/2)^2 − 9(-1/2) − 1 = 1/16 + 7/4 + 9/2 − 1 = 17/16 > 0
From the above evaluations, we can see that f(a) and f(b) have opposite signs in interval (-1/2,0) (choice A) and in interval (-1,-1/2) (choice D). Therefore, the Intermediate Value Theorem guarantees that the polynomial x^4 + 7x^2 − 9x − 1 has a root in both of these intervals.