Question

a. use the intermediate value theorem to show that the following equation has a solution on the given interval. b. use the graphing utility to find all the solutions to the equation on the given interval. c. illustrate your answers with an appropriate graph. -x³ + 6x² - 7x = 3, (-1,5) a. the intermediate value theorem states that if f is continuous on the interval and l is a number between f(a) and f(b), then number c in satisfying

Explanation:

Step1: Define the function

Let \(f(x)=-x^{3}+6x^{2}-7x - 3\). The function \(y = f(x)\) is a polynomial function, and polynomial functions are continuous everywhere, so it is continuous on the interval \((-1,5)\).

Step2: Evaluate the function at the endpoints

Calculate \(f(-1)\):
\[

$$\begin{align*} f(-1)&=-(-1)^{3}+6(-1)^{2}-7(-1)-3\\ &=1 + 6+7 - 3\\ &=11 \end{align*}$$

\]
Calculate \(f(5)\):
\[

$$\begin{align*} f(5)&=-5^{3}+6\times5^{2}-7\times5-3\\ &=-125 + 150-35 - 3\\ &=-125+150-(35 + 3)\\ &=25 - 38\\ &=-13 \end{align*}$$

\]

Step3: Apply the Intermediate - Value Theorem

Since \(f(x)\) is continuous on \([-1,5]\), and \(0\) is a number between \(f(5)=-13\) and \(f(-1)=11\), then there exists a number \(c\) in \((-1,5)\) such that \(f(c)=0\). So the equation \(-x^{3}+6x^{2}-7x = 3\) (or \(f(x)=0\)) has a solution in the interval \((-1,5)\).

Answer:

The equation \(-x^{3}+6x^{2}-7x = 3\) has a solution in the interval \((-1,5)\) as shown by the Intermediate - Value Theorem. For part b, using a graphing utility (not shown in this step - by - step solution), one can find the actual \(x\) - values where the graph of \(y=-x^{3}+6x^{2}-7x - 3\) intersects the \(x\) - axis in the interval \((-1,5)\). For part c, the graph of \(y=-x^{3}+6x^{2}-7x - 3\) would show the function crossing the \(x\) - axis within the interval \((-1,5)\).