Question

a. the intermediate value theorem states that if f is continuous on the interval a,b and l is a number strictly between f(a) and f(b), then there exists at least one number c in (a,b) satisfying f(c)=l. for which value(s) of x is the function f(x) = \sqrt{x^{4}+19x^{3}+5} continuous? a. it is continuous on 0,1, but not for all x b. it is continuous for some x, but not on 0,1 c. it is continuous for all x d. it is not continuous on any interval b. use the intermediate value theorem to show that the following equation has a solution on the given interval \sqrt{x^{4}+19x^{3}+5}=4, (0,1) c. use the graphing utility to find all the solutions to the equation on the given interval. illustrate your answers with an appropriate graph.

Explanation:

Step1: Recall continuity conditions

A function $y = f(x)=\sqrt{x^{4}+19x^{3}+5}$ is a composition of a square - root function and a polynomial function. A polynomial $p(x)=x^{4}+19x^{3}+5$ is continuous for all real - valued $x$ since $p(x)$ is of the form $a_nx^n+\cdots+a_1x + a_0$ where $n = 4$ and $a_i$ are constants. The square - root function $y=\sqrt{u}$ is continuous for $u\geq0$. For $p(x)=x^{4}+19x^{3}+5$, when $x\in[0,1]$, $p(0)=0^{4}+19\times0^{3}+5 = 5>0$ and $p(1)=1^{4}+19\times1^{3}+5=1 + 19+5=25>0$. Also, for all real $x$, $x^{4}\geq0$ and $19x^{3}+5$ makes $x^{4}+19x^{3}+5>0$ for a wide range of $x$ values. So, $f(x)$ is continuous for all $x$ because the expression inside the square - root is always positive for all real $x$.

Answer:

C. It is continuous for all x