does the intermediate value theorem guarantee a value of c in the given interval? if so, find the c - value…

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.\n40. $f(x)=\frac{x^{2}-x}{x}, f(c)=-1$ 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 $-2,2$

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.\n40. $f(x)=\frac{x^{2}-x}{x}, f(c)=-1$ 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 $-2,2$

Answer

40.

Explanation:

Step1: Simplify the function

For (f(x)=\frac{x^{2}-x}{x}), (x\neq0), and (f(x)=x - 1) for (x\neq0). 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.

Explanation:

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.

Explanation:

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\notin[-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})