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Multiple Choice
6) Use the Intermediate Value Theorem to show that the equation has a solution on the interval . Which of the following justifies the existence of a solution?
A
The function is increasing on .
B
The function has a maximum on .
C
The function is differentiable on , and .
D
The function is continuous on , and and have opposite signs.
Verified step by step guidance
1
Step 1: Recall the Intermediate Value Theorem (IVT). It states that if a function f(x) is continuous on a closed interval [a, b], and f(a) and f(b) have opposite signs, then there exists at least one c in (a, b) such that f(c) = 0.
Step 2: Define the function f(x) = x^2 - 6x - 3. Verify that f(x) is a polynomial function, which is continuous everywhere, including on the interval [0, 7].
Step 3: Evaluate f(x) at the endpoints of the interval. Compute f(0) and f(7): f(0) = (0)^2 - 6(0) - 3 and f(7) = (7)^2 - 6(7) - 3. Determine the signs of f(0) and f(7).
Step 4: Check if f(0) and f(7) have opposite signs. If f(0) < 0 and f(7) > 0 (or vice versa), then the Intermediate Value Theorem guarantees that there is at least one solution to f(x) = 0 in the interval [0, 7].
Step 5: Conclude that the correct justification for the existence of a solution is: 'The function f(x) = x^2 - 6x - 3 is continuous on [0, 7], and f(0) and f(7) have opposite signs.'