6) Use the Intermediate Value Theorem to show that the equation $x^2-6x-3=0$ has a solution on the interval $[0,7]$. Which of the following justifies the existence of a solution?

• A. The function $f\left(x\right)=x^2-6x-3$ is increasing on $[0,7]$.
• B. The function $f\left(x\right)=x^2-6x-3$ has a maximum on $[0,7]$.
• C. The function $f\left(x\right)=x^2-6x-3$ is differentiable on $[0,7]$, and $f\left(0\right)=f\left(7\right)$.
• D. The function $f\left(x\right)=x^2-6x-3$ is continuous on $[0,7]$, and $f\left(0\right)$ and $f\left(7\right)$ have opposite signs.

Answer: (D)

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.
2. 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].
3. 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).
4. 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].
5. 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.'