Question

using the intermediate value theorem in exercises 95–100, verify that the intermediate value theorem applies to the indicated interval and find the value of c guaranteed by the theorem. 95. ( f(x) = x^2 + x - 1 ), (0, 5), ( f(c) = 11 ) 96. ( f(x) = x^2 - 6x + 8 ), (0, 3), ( f(c) = 0 ) 97. ( f(x) = sqrt{x + 5} - 2 ), (0, 5), ( f(c) = 1 ) 98. ( f(x) = sqrt3{x} + 8 ), (-9, -6), ( f(c) = 6 ) 99. ( f(x) = \frac{x - x^3}{x - 4} ), (1, 3), ( f(c) = 3 ) 100. ( f(x) = \frac{x^2 + x}{x - 1} ), (left \frac{5}{2}, 4
ight), ( f(c) = 6 )

Explanation:

Let's solve problem 95: \( f(x) = x^3 + x - 1 \), interval \([0, 5]\), \( f(c) = 11 \) using the Intermediate Value Theorem (IVT).

The Intermediate Value Theorem states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and \( k \) is a number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \) in \([a, b]\) such that \( f(c) = k \).

Step 1: Check Continuity

The function \( f(x) = x^3 + x - 1 \) is a polynomial function. Polynomial functions are continuous everywhere, so \( f(x) \) is continuous on \([0, 5]\).

Step 2: Evaluate \( f(0) \) and \( f(5) \)

• Calculate \( f(0) \):

\( f(0) = 0^3 + 0 - 1 = -1 \)

• Calculate \( f(5) \):

\( f(5) = 5^3 + 5 - 1 = 125 + 5 - 1 = 129 \)

Step 3: Check if \( k = 11 \) is between \( f(0) \) and \( f(5) \)

We have \( f(0) = -1 \) and \( f(5) = 129 \). Since \( -1 < 11 < 129 \), by IVT, there exists a \( c \in [0, 5] \) such that \( f(c) = 11 \).

Step 4: Solve \( f(c) = 11 \) for \( c \)

Set \( f(c) = 11 \):
\( c^3 + c - 1 = 11 \)
\( c^3 + c - 12 = 0 \)

We can try to find a root by trial. Let's test \( c = 2 \):
\( 2^3 + 2 - 12 = 8 + 2 - 12 = -2 \) (not zero)

Test \( c = 3 \):
\( 3^3 + 3 - 12 = 27 + 3 - 12 = 18 \) (not zero)

Wait, maybe I made a mistake. Wait, \( f(2) = 8 + 2 - 1 = 9 \), \( f(3) = 27 + 3 - 1 = 29 \). Wait, \( k = 11 \) is between \( f(2) = 9 \) and \( f(3) = 29 \). Let's try \( c = 2 \): \( f(2) = 8 + 2 - 1 = 9 \), \( c = 2.1 \): \( 2.1^3 + 2.1 - 1 = 9.261 + 2.1 - 1 = 10.361 \), \( c = 2.2 \): \( 2.2^3 + 2.2 - 1 = 10.648 + 2.2 - 1 = 11.848 \). Wait, maybe there's a calculation error. Wait, the original equation: \( c^3 + c - 1 = 11 \) => \( c^3 + c - 12 = 0 \). Let's use rational root theorem. Possible rational roots are factors of 12 over factors of 1: ±1, ±2, ±3, ±4, ±6, ±12.

Test \( c = 2 \): \( 8 + 2 - 12 = -2 \)

Test \( c = 3 \): \( 27 + 3 - 12 = 18 \)

Test \( c = \sqrt[3]{12 - c} \)? Wait, maybe I made a mistake in the problem. Wait, the problem says \( f(x) = x^3 + x - 1 \), \( f(c) = 11 \). Let's re-express:

\( c^3 + c - 1 = 11 \) => \( c^3 + c - 12 = 0 \)

Let's use the Newton-Raphson method. Let's take \( x_0 = 2 \), \( f(2) = 8 + 2 - 12 = -2 \), \( f'(x) = 3x^2 + 1 \), \( f'(2) = 13 \)

Next iteration: \( x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 2 - \frac{-2}{13} = 2 + \frac{2}{13} \approx 2.1538 \)

\( f(2.1538) \approx (2.1538)^3 + 2.1538 - 12 \approx 10.07 + 2.1538 - 12 \approx 0.2238 \)

Next iteration: \( x_2 = 2.1538 - \frac{0.2238}{3*(2.1538)^2 + 1} \approx 2.1538 - \frac{0.2238}{14.28 + 1} \approx 2.1538 - 0.0147 \approx 2.1391 \)

\( f(2.1391) \approx (2.1391)^3 + 2.1391 - 12 \approx 9.83 + 2.1391 - 12 \approx -0.0309 \)

Next iteration: \( x_3 = 2.1391 - \frac{-0.0309}{3*(2.1391)^2 + 1} \approx 2.1391 + \frac{0.0309}{14.04 + 1} \approx 2.1391 + 0.00205 \approx 2.14115 \)

\( f(2.14115) \approx (2.14115)^3 + 2.14115 - 12 \approx 9.89 + 2.14115 - 12 \approx 0.03115 \)? Wait, maybe my initial approach is wrong. Wait, maybe the problem is \( f(x) = x^2 + x - 1 \)? Let's check. If \( f(x) = x^2 + x - 1 \), then \( f(3) = 9 + 3 - 1 = 11 \). Ah! Maybe it's a typo, \( x^2 \) instead of \( x^3 \). Let's assume that (since \( x^3 \) gives a non-integer, but \( x^2 \) gives \( c = 3 \)).

If \( f(x) = x^2 + x - 1 \), then:

\( f(0) = 0 + 0 - 1 = -1 \)

\( f(5) = 25 + 5 - 1 = 29 \)

\( 11 \) is between \( -1 \) and \( 29 \). Then \( f(c) = 11 \) => \( c^2 + c - 1 = 11 \) => \( c^2 + c - 12 = 0 \) => \( (c + 4)(c - 3) = 0 \) => \( c = 3 \) (since \( c \in [0,5] \), \( c = -4 \) is excluded).

So probably a typo, \( x^2 \) instead of \( x^3 \). So \( c = 3 \).

Answer:

\( c = 3 \) (assuming \( f(x) = x^2 + x - 1 \) due to the integer solution; if \( f(x) = x^3 + x - 1 \), the solution is approximately \( 2.14 \), but likely a typo).