The problem asks to find the real zero of the function $f(x) = 3x^5 + 7x - 4$ using the Intermediate Value Theorem, correct to two decimal places.

Solution:

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 (i.e., one is positive and the other is negative), then there is at least one root (zero) between $a$ and $b$.

Step-by-Step Process:

1. Select an interval where the function changes sign.
2. Evaluate $f(x)$ at the endpoints of the interval.
3. Use bisection to repeatedly narrow down the interval until the root is accurate to two decimal places.

Let’s start by checking different values of $x$ to find such an interval.

Let me calculate the function values for different $x$ values and then narrow down the root interval.

Trying different values of $x$:

• $f(0) = 3(0)^5 + 7(0) - 4 = -4$
• $f(1) = 3(1)^5 + 7(1) - 4 = 6$
• $f(-1) = 3(-1)^5 + 7(-1) - 4 = -14$
• $f(0.5) = 3(0.5)^5 + 7(0.5) - 4 = 0.8125$

We can see that the function changes sign between $x = 0$ and $x = 1$, so we can apply the IVT to the interval $[0, 1]$.

Now, we perform the bisection method between $x = 0$ and $x = 1$.

Let me compute it step by step.The real zero of the function $f(x) = 3x^5 + 7x - 4$, using the Intermediate Value Theorem and the bisection method, is approximately $x = 0.5499$.

Would you like any further explanation or details on this process?

Here are some related questions for deeper understanding:

1. How does the Intermediate Value Theorem ensure the existence of a root?
2. Can the bisection method fail, and if so, under what circumstances?
3. How would you apply the bisection method to more complex functions?
4. Why is it important to round intermediate values in numerical methods?
5. What other numerical methods can you use to find roots of functions?

Tip: When using the bisection method, the initial interval choice is critical—ensure the function changes sign between the endpoints!