Step 1: Verify that the IVT applies

The Intermediate Value Theorem (IVT) 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 $f(c) = 0$.

To apply the IVT for the function $f(x) = 2x \cos(x) - 3x$ on the interval $[0, 1]$, we need to:

1. Check the continuity of $f(x)$ on $[0, 1]$.
2. Evaluate $f(0)$ and $f(1)$.
3. Confirm that $f(0)$ and $f(1)$ have opposite signs.

Step 2: Check the continuity of $f(x)$

The function $f(x) = 2x \cos(x) - 3x$ involves basic operations (multiplication, subtraction) of continuous functions (since both $x$ and $\cos(x)$ are continuous everywhere). Therefore, $f(x)$ is continuous on the interval $[0, 1]$.

Step 3: Evaluate $f(0)$ and $f(1)$

• At $x = 0$: $f(0) = 2(0) \cos(0) - 3(0) = 0$

• At $x = 1$: $f(1) = 2(1) \cos(1) - 3(1) = 2 \cos(1) - 3 \approx 2(0.5403) - 3 \approx 1.0806 - 3 = -1.9194$

Step 4: Apply the IVT

• $f(0) = 0$
• $f(1) \approx -1.9194$

Since $f(0) = 0$, we already know that the function has a zero at $x = 0$. Therefore, the IVT does not give us additional information on this particular interval, but since $f(0) = 0$, we can confidently say that $x = 0$ is the zero of the function.

Conclusion:

By the Intermediate Value Theorem, the function does have a zero in the interval $[0, 1]$, specifically at $x = 0$.

Would you like further details or have additional questions? Here are some related questions to explore:

1. What is the behavior of the function as $x$ increases from 0 to 1?
2. How can you approximate zeros using numerical methods, like the bisection method?
3. How would the IVT apply if the signs of $f(0)$ and $f(1)$ were different?
4. What would happen if the function had a discontinuity at some point in the interval?
5. How does the IVT work when dealing with more complex functions, such as trigonometric polynomials?

Tip: Always check whether the function is continuous on the interval in question when applying the IVT!