use the intermediate value theorem to prove that the equation x³ = -5 has at least one negative solution. in…

Explanation:

Step1: Define the function

Let $f(x)=x^{3}+ 5$. The function $y = f(x)$ is a polynomial function, so it is continuous everywhere.

Step2: Evaluate the function at the endpoints of each interval

For interval A. $[1,2]$:

$f(1)=1^{3}+5=6$ and $f(2)=2^{3}+5 = 8 + 5=13$. Since $f(1)>0$ and $f(2)>0$, there is no sign - change in this interval.

For interval B. $[0,1]$:

$f(0)=0^{3}+5 = 5$ and $f(1)=1^{3}+5=6$. Since $f(0)>0$ and $f(1)>0$, there is no sign - change in this interval.

For interval C. $[-2,-1]$:

$f(-2)=(-2)^{3}+5=-8 + 5=-3$ and $f(-1)=(-1)^{3}+5=-1 + 5 = 4$. Since $f(-2)<0$ and $f(-1)>0$, by the Intermediate Value Theorem, there exists at least one $c\in[-2,-1]$ such that $f(c)=0$, which means $c^{3}=-5$.

For interval D. $[-1,0]$:

$f(-1)=(-1)^{3}+5 = 4$ and $f(0)=0^{3}+5=5$. Since $f(-1)>0$ and $f(0)>0$, there is no sign - change in this interval.

Answer:

C. $[-2,-1]$