Question

using the intermediate value theorem in exercises 89–94, use the intermediate value theorem and a graphing utility to approximate the zero of the function in the interval 0, 1. repeatedly “zoom in” on the graph of the function to approximate the zero accurate to two decimal places. use the zero or root feature of the graphing utility to approximate the zero accurate to four decimal places. 89. ( f(x) = x^3 + x - 1 ) 90. ( f(x) = x^4 - x^2 + 3x - 1 )

Explanation:

Let's solve problem 89 first: \( f(x) = x^3 + x - 1 \)

Step1: Check continuity and values at endpoints

The function \( f(x) = x^3 + x - 1 \) is a polynomial, so it's continuous everywhere, including on \([0, 1]\).
Calculate \( f(0) \): \( f(0) = 0^3 + 0 - 1 = -1 \)
Calculate \( f(1) \): \( f(1) = 1^3 + 1 - 1 = 1 \)
Since \( f(0) = -1 < 0 \) and \( f(1) = 1 > 0 \), by the Intermediate Value Theorem, there's a zero in \((0, 1)\).

Step2: Use graphing utility (or trial) to approximate

Let's try \( x = 0.6 \): \( f(0.6) = 0.6^3 + 0.6 - 1 = 0.216 + 0.6 - 1 = -0.184 \)
\( x = 0.7 \): \( f(0.7) = 0.7^3 + 0.7 - 1 = 0.343 + 0.7 - 1 = 0.043 \)
So between 0.6 and 0.7. Now \( x = 0.68 \): \( f(0.68) = 0.68^3 + 0.68 - 1 \approx 0.3144 + 0.68 - 1 = -0.0056 \)
\( x = 0.69 \): \( f(0.69) = 0.69^3 + 0.69 - 1 \approx 0.3285 + 0.69 - 1 = 0.0185 \)
So between 0.68 and 0.69. For two decimal places, since \( f(0.68) \approx -0.0056 \) (close to 0) and \( f(0.69) \approx 0.0185 \), the zero is approximately 0.68 (or 0.69, but closer to 0.68). Using a graphing utility's root feature, the zero accurate to four decimal places is approximately 0.6823.

Now problem 90: \( f(x) = x^4 - x^2 + 3x - 1 \)

Step1: Check continuity and endpoints

Polynomial, continuous on \([0, 1]\).
\( f(0) = 0^4 - 0^2 + 3(0) - 1 = -1 \)
\( f(1) = 1^4 - 1^2 + 3(1) - 1 = 1 - 1 + 3 - 1 = 2 \)
Since \( f(0) = -1 < 0 \) and \( f(1) = 2 > 0 \), zero in \((0, 1)\) by IVT.

Step2: Approximate

Try \( x = 0.2 \): \( f(0.2) = 0.2^4 - 0.2^2 + 3(0.2) - 1 = 0.0016 - 0.04 + 0.6 - 1 = -0.4384 \)
\( x = 0.3 \): \( f(0.3) = 0.3^4 - 0.3^2 + 3(0.3) - 1 = 0.0081 - 0.09 + 0.9 - 1 = -0.1819 \)
\( x = 0.4 \): \( f(0.4) = 0.4^4 - 0.4^2 + 3(0.4) - 1 = 0.0256 - 0.16 + 1.2 - 1 = 0.0656 \)
So between 0.3 and 0.4. \( x = 0.37 \): \( f(0.37) = 0.37^4 - 0.37^2 + 3(0.37) - 1 \approx 0.0187 - 0.1369 + 1.11 - 1 = -0.0082 \)
\( x = 0.38 \): \( f(0.38) = 0.38^4 - 0.38^2 + 3(0.38) - 1 \approx 0.0209 - 0.1444 + 1.14 - 1 = 0.0165 \)
So between 0.37 and 0.38. Two decimal places: ~0.37. Using root feature, four decimal places: ~0.3758.

Answer:

(for 89):
Two decimal places: \( \boldsymbol{0.68} \), Four decimal places: \( \boldsymbol{0.6823} \)