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
x^3 - 5x^2 + 2x = - 2, (-1,5)
evaluate the function f(x) at the right endpoint
the value of the function at the right endpoint is 10
(type an integer or decimal rounded to three decimal places as needed.)
why can the intermediate value theorem be used to show that the equation has a solution on (-1,5)?
a. it can be used because f(x)=x^3 - 5x^2 + 2x is defined on (-1,5) and -2 < f(-1) < f(5)
b. it can be used because f(x)=x^3 - 5x^2 + 2x is continuous on -1,5 and the function is defined at x = - 1 and x = 5
c. it can be used because f(x)=x^3 - 5x^2 + 2x is defined on (-1,5) and f(-1) < f(5) < -2
d. it can be used because f(x)=x^3 - 5x^2 + 2x is continuous on -1,5 and -2 is between f(-1) and f(5)
b. there is/are a solution(s) to the equation in (-1,5) at x ≈
(type integer or decimals rounded to three decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Recall Intermediate Value Theorem

The Intermediate Value Theorem states that if a function $y = f(x)$ 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 the open - interval $(a,b)$ such that $f(c)=k$. For the function $f(x)=x^{3}-5x^{2}+2x$, it is a polynomial function, and polynomial functions are continuous everywhere. So, $f(x)$ is continuous on $[-1,5]$. We want to show that the equation $x^{3}-5x^{2}+2x=-2$ has a solution in $(-1,5)$. Let $k = - 2$. We need to find $f(-1)$ and $f(5)$.

Step2: Calculate $f(-1)$

\[

$$\begin{align*} f(-1)&=(-1)^{3}-5(-1)^{2}+2(-1)\\ &=-1 - 5-2\\ &=-8 \end{align*}$$

\]

Step3: Calculate $f(5)$

\[

$$\begin{align*} f(5)&=5^{3}-5\times5^{2}+2\times5\\ &=125-125 + 10\\ &=10 \end{align*}$$

\]
Since $-2$ is between $f(-1)=-8$ and $f(5)=10$ and $f(x)$ is continuous on $[-1,5]$, by the Intermediate Value Theorem, the equation $x^{3}-5x^{2}+2x=-2$ has a solution in $(-1,5)$.

Step4: Use a graphing utility

To find the solutions of the equation $x^{3}-5x^{2}+2x+2 = 0$ in the interval $(-1,5)$ using a graphing utility (such as a graphing calculator or software like Desmos), we graph the function $y=x^{3}-5x^{2}+2x + 2$. The $x$ - values of the points where the graph of $y=x^{3}-5x^{2}+2x + 2$ intersects the $x$ - axis in the interval $(-1,5)$ are the solutions.
By using a graphing utility, the solutions of the equation $x^{3}-5x^{2}+2x=-2$ (or $x^{3}-5x^{2}+2x + 2=0$) in the interval $(-1,5)$ are $x\approx0.438,4.562$.

Answer:

a. The correct option is D.
b. $0.438,4.562$