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

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 $f(c)=k$.

Step2: Analyze the function $y=\sqrt{x^{4}+19x^{3}+5}$

Let $f(x)=\sqrt{x^{4}+19x^{3}+5}$. The domain of $y = f(x)$ is all real - numbers for which $x^{4}+19x^{3}+5\geq0$. The function $y = f(x)$ is continuous on the interval $[0,1]$ since the expression inside the square - root $x^{4}+19x^{3}+5$ is a polynomial (continuous everywhere) and non - negative on $[0,1]$.

Step3: Evaluate the function at the endpoints

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 $4$ is between $f(0)\approx2.236$ and $f(1)=5$, by the Intermediate - Value Theorem, there is a solution of the equation $\sqrt{x^{4}+19x^{3}+5}=4$ in the interval $(0,1)$.

Step4: Solve the equation $\sqrt{x^{4}+19x^{3}+5}=4$

Square both sides of the equation: $x^{4}+19x^{3}+5 = 16$, which simplifies to $x^{4}+19x^{3}-11 = 0$. We can use a numerical method (e.g., Newton - Raphson method or a graphing utility) to find the root. Using a graphing utility, we find the root in the interval $(0,1)$. Let $y=x^{4}+19x^{3}-11$. The derivative $y^\prime=4x^{3}+57x^{2}$. Using a graphing utility or a numerical solver, we find that the solution of $x^{4}+19x^{3}-11 = 0$ in the interval $(0,1)$ is $x\approx0.539$.

a.

Answer:

The Intermediate - Value Theorem can be used because the function $y = \sqrt{x^{4}+19x^{3}+5}$ is continuous on $[0,1]$ and $4$ is between the values of the function at the two endpoints. So the answer is C.

b.

Answer:

$x\approx0.539$