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.
√(x⁴ + 19x³ + 5)=4, (0,1)
(type an integer or decimal rounded to three decimal places as needed )

b. the value of the function at the right - endpoint is undefined

why can the intermediate value theorem be used to show that the equation has a solution on (0,1)?

a. it can be used because √(x⁴ + 19x³ + 5) is continuous on 0,1 and the function is defined at x = 0 and x = 1
b. it can be used because √(x⁴ + 19x³ + 5) is defined on (0,1) and 4 is less than the values of the function at the two endpoints
c. it can be used because √(x⁴ + 19x³ + 5) is continuous on 0,1 and 4 lies between the values of the function at the two endpoints.
d. it can be used because √(x⁴ + 19x³ + 5) is defined on (0,1) and 4 is greater than the values of the function at the two endpoints

b. there is a solution to the equation on (0,1) at x≈
(round to the nearest thousandth as needed )

Explanation:

Step1: Evaluate function at endpoints

Let $f(x)=\sqrt{x^{4}+19x^{3}+5}$. When $x = 0$, $f(0)=\sqrt{0^{4}+19\times0^{3}+5}=\sqrt{5}\approx2.236$. When $x = 1$, $f(1)=\sqrt{1^{4}+19\times1^{3}+5}=\sqrt{1 + 19+5}=\sqrt{25}=5$. Since $y = f(x)$ is a square - root function and the expression inside the square - root $x^{4}+19x^{3}+5$ is a polynomial (so continuous everywhere), $y = f(x)$ is continuous on $[0,1]$. And $2.236<4<5$, by the Intermediate Value Theorem, there exists a $c\in(0,1)$ such that $f(c)=4$.

Step2: Use graphing utility to find solution

Using a graphing utility (such as a graphing calculator or software like Desmos), we graph $y=\sqrt{x^{4}+19x^{3}+5}$ and $y = 4$ and find the $x$ - value of their intersection point in the interval $(0,1)$. Let $y_1=\sqrt{x^{4}+19x^{3}+5}$ and $y_2 = 4$. We set $y_1=y_2$, so $x^{4}+19x^{3}+5 = 16$ (squaring both sides, note we need to check for extraneous solutions later, but in this context it's valid as the original function is non - negative). Let $g(x)=x^{4}+19x^{3}-11$. Using a numerical method (such as Newton - Raphson method or the zero - finding feature on a graphing utility), we find that $x\approx0.539$.

Answer:

a. C. It can be used because $\sqrt{x^{4}+19x^{3}+5}$ is continuous on $[0,1]$ and $4$ lies between the values of the function at the two endpoints.
b. $0.539$