To solve this problem using the Intermediate Value Theorem (IVT), we need to verify that the function changes signs over the given interval. The IVT states that if a continuous function $f(x)$ takes on opposite signs at two points $a$ and $b$, i.e., $f(a)f(b) < 0$, then there is at least one root in the interval $[a,b]$.

Let's analyze each part.

(a) $25 - 8x^2 - x^3 = 0$ on $[-2, 4]$

1. Define the function: $f(x) = 25 - 8x^2 - x^3$.
2. Compute $f(-2)$ and $f(4)$: $f(-2) = 25 - 8(-2)^2 - (-2)^3 = 25 - 8(4) + 8 = 25 - 32 + 8 = 1$ $f(4) = 25 - 8(4)^2 - (4)^3 = 25 - 8(16) - 64 = 25 - 128 - 64 = -167$
3. Since $f(-2) = 1$ and $f(4) = -167$, the function changes signs, so by the IVT, there is at least one solution in $[-2, 4]$.

(b) $w^2 - 4\ln(5w + 2) = 0$ on $[0, 4]$

1. Define the function: $f(w) = w^2 - 4\ln(5w + 2)$.
2. Compute $f(0)$ and $f(4)$: $f(0) = 0^2 - 4\ln(5(0) + 2) = -4\ln(2) \approx -2.77$ $f(4) = 4^2 - 4\ln(5(4) + 2) = 16 - 4\ln(22) \approx 16 - 12.18 = 3.82$
3. Since $f(0) \approx -2.77$ and $f(4) \approx 3.82$, the function changes signs, so by the IVT, there is at least one solution in $[0, 4]$.

(c) $4t + 10e^t = e^{2t}$ on $[1, 3]$

1. Define the function: $f(t) = 4t + 10e^t - e^{2t}$.
2. Compute $f(1)$ and $f(3)$: $f(1) = 4(1) + 10e^1 - e^{2(1)} = 4 + 10e - e^2 \approx 4 + 27.18 - 7.39 = 23.79$ $f(3) = 4(3) + 10e^3 - e^{2(3)} = 12 + 10e^3 - e^6 \approx 12 + 201.72 - 403.43 = -189.71$
3. Since $f(1) \approx 23.79$ and $f(3) \approx -189.71$, the function changes signs, so by the IVT, there is at least one solution in $[1, 3]$.

(d) $\frac{t^2 + t - 15}{t - 8} = 0$ on $[-5, 1]$

1. Define the function: $f(t) = \frac{t^2 + t - 15}{t - 8}$.
2. Compute $f(-5)$ and $f(1)$: $f(-5) = \frac{(-5)^2 + (-5) - 15}{-5 - 8} = \frac{25 - 5 - 15}{-13} = \frac{5}{-13} \approx -0.38$ $f(1) = \frac{(1)^2 + 1 - 15}{1 - 8} = \frac{1 + 1 - 15}{-7} = \frac{-13}{-7} = 1.86$
3. Since $f(-5) \approx -0.38$ and $f(1) \approx 1.86$, the function changes signs, so by the IVT, there is at least one solution in $[-5, 1]$.

Summary

• By the Intermediate Value Theorem, each equation has at least one solution in the given interval since the function values at the endpoints of each interval have opposite signs.

Do you have any questions or would you like more details?

Here are 5 relative questions:

1. How is the Intermediate Value Theorem useful in root-finding methods like bisection?
2. Can a function have multiple solutions in an interval where the IVT applies?
3. How does continuity affect the application of the IVT?
4. What happens if the function does not change sign on the interval?
5. Can the IVT be applied to functions with vertical asymptotes?

Tip: Always check if the function is continuous on the interval before applying the IVT.