Question

does the intermediate value theorem guarantee a value of c in the given interval? if so, find the c - value. if not, explain why not.

1. $f(x)=\frac{x^{2}-x}{x}, f(c)=-1$ on $-2,2$ 41. $f(x)=x^{2}-x, f(c)=-1$ on $-2,2$ 42. $f(x)=x^{2}-x, f(c)=5$ on $-2,2$

Explanation:

40.

Step1: Simplify the function

For \(f(x)=\frac{x^{2}-x}{x}\), \(x
eq0\), and \(f(x)=x - 1\) for \(x
eq0\).
The function \(y = f(x)\) is not continuous on the closed - interval \([-2,2]\) because it is not defined at \(x = 0\). The Intermediate - Value Theorem requires the function to be continuous on the closed interval \([a,b]\). So the Intermediate - Value Theorem does not guarantee a value of \(c\) in the interval \([-2,2]\) for \(f(c)=-1\).

41.

Step1: Check continuity

The function \(f(x)=x^{2}-x\) is a polynomial function. Polynomial functions are continuous everywhere, so \(f(x)\) is continuous on the closed interval \([-2,2]\).

Step2: Evaluate the function at the endpoints

Calculate \(f(-2)=(-2)^{2}-(-2)=4 + 2=6\) and \(f(2)=2^{2}-2=4 - 2 = 2\).
We want to find \(c\) such that \(f(c)=-1\). But \(-1\) is not between \(f(-2)=6\) and \(f(2)=2\). So the Intermediate - Value Theorem does not guarantee a value of \(c\) in the interval \([-2,2]\) for \(f(c)=-1\).

42.

Step1: Check continuity

The function \(f(x)=x^{2}-x\) is a polynomial function, so it is continuous on the closed interval \([-2,2]\).

Step2: Evaluate the function at the endpoints

Calculate \(f(-2)=(-2)^{2}-(-2)=4 + 2=6\) and \(f(2)=2^{2}-2=4 - 2 = 2\).
We want to find \(c\) such that \(f(c)=5\). Since \(2<5<6\), by the Intermediate - Value Theorem, there exists a \(c\in[-2,2]\) such that \(f(c)=5\).
Set \(f(c)=c^{2}-c = 5\), then \(c^{2}-c - 5=0\).

Step3: Solve the quadratic equation

Using the quadratic formula \(c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b=-1\), and \(c=-5\).
\[c=\frac{1\pm\sqrt{(-1)^{2}-4\times1\times(-5)}}{2\times1}=\frac{1\pm\sqrt{1 + 20}}{2}=\frac{1\pm\sqrt{21}}{2}\]
\(c=\frac{1+\sqrt{21}}{2}\approx\frac{1 + 4.58}{2}=\frac{5.58}{2}=2.79
otin[-2,2]\) and \(c=\frac{1-\sqrt{21}}{2}\approx\frac{1-4.58}{2}=\frac{-3.58}{2}=-1.79\in[-2,2]\)

Answer:

1. The Intermediate - Value Theorem does not guarantee a value of \(c\) because the function \(f(x)=\frac{x^{2}-x}{x}\) is not continuous on \([-2,2]\).
2. The Intermediate - Value Theorem does not guarantee a value of \(c\) because \(-1\) is not between \(f(-2) = 6\) and \(f(2)=2\).
3. Yes, \(c=\frac{1-\sqrt{21}}{2}\)