Question

a. use the intermediate value theorem to show that the following equation has a solution on the given interval 4x³ +x +3=0. (-1,1)
b. use a graphing utility to find all the solutions to the equation on the given interval
c. illustrate your answers with an appropriate graph.
a. the intermediate value theorem states that if f is continuous on the interval a,b and l is a number strictly between f(a) and f(b), then there exists at least one number c in (a,b) satisfying f(c)=l. for which values of x is the function f(x)=4x³ +x +3 continuous?
options:
a it is continuous for some x, but not on -1,1.
b it is continuous on -1,1, but not for all x
c it is continuous for all x
d it is not continuous on any interval
evaluate the function f(x) at the left endpoint.
( type an integer or decimal rounded to three decimal places as needed )
the value of the function at the left endpoint is -2.
evaluate the function f(x) at the right endpoint.
( type an integer or decimal rounded to three decimal places as needed )
the value of the function at the right endpoint is 8.
why can the intermediate value theorem be used to show that the equation has a solution on (-1,1)?
options:
a it can be used because f(x)=4x³ +x +3 is continuous on -1,1 and the function is defined at x=-1 and x=1
b it can be used because f(x)=4x³ +x +3 is defined on (-1,1) and 0 < f(-1) < f(1)
c it can be used because f(x)=4x³ +x +3 is defined on (-1,1) and f(-1) < f(1) < 0
d it can be used because f(x)=4x³ +x +3 is continuous on -1,1 and 0 is between f(-1) and f(1)
b. there is/are a solution(s) to the equation in (-1,1) at x≈
(type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Part a

Step1: Recall continuity of polynomials

Polynomial functions (like \( f(x) = 4x^3 + x + 3 \)) are continuous everywhere, i.e., for all \( x \) in \( \mathbb{R} \). So we analyze the options:

• Option A: Says continuous for some \( x \), not on \([-1,1]\) – incorrect, as polynomials are continuous on entire intervals.
• Option B: Says continuous on \([-1,1]\) but not for all \( x \) – incorrect, since it's continuous everywhere.
• Option C: Says continuous for all \( x \) – correct, as \( f(x) \) is a polynomial.
• Option D: Says not continuous on any interval – incorrect, polynomials are continuous on all intervals.

Step2: Evaluate left endpoint (\( x = -1 \))

Substitute \( x = -1 \) into \( f(x) \):
\( f(-1) = 4(-1)^3 + (-1) + 3 = 4(-1) - 1 + 3 = -4 - 1 + 3 = -2 \).

Step3: Evaluate right endpoint (\( x = 1 \))

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = 4(1)^3 + 1 + 3 = 4 + 1 + 3 = 8 \).

Step4: Apply Intermediate Value Theorem (IVT) logic

IVT applies when \( f \) is continuous on \([a,b]\) and \( L \) is between \( f(a) \) and \( f(b) \). Here, \( f(-1) = -2 \), \( f(1) = 8 \), and \( 0 \) is between \(-2\) and \( 8 \) (since \( -2 < 0 < 8 \)). So we check the options for why IVT applies:

• Option A: Mentions defined at endpoints, but IVT needs continuity and \( 0 \) between \( f(-1) \) and \( f(1) \) – incorrect.
• Option B: Says \( 0 < f(-1) < f(1) \), but \( f(-1) = -2 < 0 \) – incorrect.
• Option C: Says \( f(-1) < f(1) < 0 \), but \( f(1) = 8 > 0 \) – incorrect.
• Option D: Says \( f \) is continuous on \([-1,1]\) and \( 0 \) is between \( f(-1) \) and \( f(1) \) – correct, since \( -2 < 0 < 8 \) and \( f \) is continuous.

Part b

Step1: Set up the equation

We need to solve \( 4x^3 + x + 3 = 0 \) for \( x \) in \((-1,1)\). Let \( f(x) = 4x^3 + x + 3 \). We know \( f(-1) = -2 \) and \( f(1) = 8 \), and \( f \) is continuous. We can use the Intermediate Value Theorem or numerical methods (like trial and error with decimals) to approximate \( x \).

Step2: Test \( x = -0.9 \)

\( f(-0.9) = 4(-0.9)^3 + (-0.9) + 3 = 4(-0.729) - 0.9 + 3 = -2.916 - 0.9 + 3 = -0.816 \).

Step3: Test \( x = -0.8 \)

\( f(-0.8) = 4(-0.8)^3 + (-0.8) + 3 = 4(-0.512) - 0.8 + 3 = -2.048 - 0.8 + 3 = 0.152 \).

Step4: Use linear approximation (or more trials)

Since \( f(-0.9) \approx -0.816 \) and \( f(-0.8) \approx 0.152 \), the root is between \(-0.9\) and \(-0.8\). Let's use linear interpolation. The difference in \( x \): \( -0.8 - (-0.9) = 0.1 \). The difference in \( f(x) \): \( 0.152 - (-0.816) = 0.968 \). We need to find \( \Delta x \) where \( -0.816 + 0.968 \cdot \frac{\Delta x}{0.1} = 0 \). Solving: \( 0.968 \cdot \frac{\Delta x}{0.1} = 0.816 \implies \Delta x \approx \frac{0.816 \cdot 0.1}{0.968} \approx 0.0843 \). So \( x \approx -0.9 + 0.0843 \approx -0.8157 \approx -0.816 \) (rounded to three decimals). Testing \( x = -0.816 \):
\( f(-0.816) = 4(-0.816)^3 + (-0.816) + 3 \approx 4(-0.547) - 0.816 + 3 \approx -2.188 - 0.816 + 3 \approx -0.004 \approx 0 \) (close enough).

Answer:

s:

Part a

• For continuity: \(\boldsymbol{\text{Option C}}\) (It is continuous for all \( x \))
• Left endpoint: \(\boldsymbol{-2}\)
• Right endpoint: \(\boldsymbol{8}\)
• IVT reason: \(\boldsymbol{\text{Option D}}\) (It can be used because \( f(x) = 4x^3 + x + 3 \) is continuous on \([-1,1]\) and \( 0 \) is between \( f(-1) \) and \( f(1) \))

Part b

Solution to \( 4x^3 + x + 3 = 0 \) in \((-1,1)\): \(\boldsymbol{-0.816}\) (rounded to three decimal places)