a. use the intermediate value theorem to show that the following equation has a solution on the given…

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

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

Answer

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)).