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Let 
h(x)=(1)/(x).
Can we use the intermediate value theorem to say the equation 
h(x)=0.5 has a solution where 
-1 <= x <= 1 ?
Choose 1 answer:
(A) No, since the function is not continuous on that interval.
(B) No, since 0.5 is not between 
h(-1) and 
h(1).
(C) Yes, both conditions for using the intermediate value theorem have been met.

Let h(x)=1x h(x)=\frac{1}{x} .\newlineCan we use the intermediate value theorem to say the equation h(x)=0.5 h(x)=0.5 has a solution where 1x1 -1 \leq x \leq 1 ?\newlineChoose 11 answer:\newline(A) No, since the function is not continuous on that interval.\newline(B) No, since 00.55 is not between h(1) h(-1) and h(1) h(1) .\newline(C) Yes, both conditions for using the intermediate value theorem have been met.

Full solution

Q. Let h(x)=1x h(x)=\frac{1}{x} .\newlineCan we use the intermediate value theorem to say the equation h(x)=0.5 h(x)=0.5 has a solution where 1x1 -1 \leq x \leq 1 ?\newlineChoose 11 answer:\newline(A) No, since the function is not continuous on that interval.\newline(B) No, since 00.55 is not between h(1) h(-1) and h(1) h(1) .\newline(C) Yes, both conditions for using the intermediate value theorem have been met.
  1. Check Conditions: To use the intermediate value theorem, we need to check two conditions: the function must be continuous on the closed interval a,ba, b, and the value we are looking for (in this case, 0.50.5) must be between the values of the function at the endpoints of the interval.
  2. Function Continuity: First, let's check if the function h(x)=1xh(x) = \frac{1}{x} is continuous on the interval [1,1][-1, 1]. We know that the function 1x\frac{1}{x} is not defined at x=0x = 0, which lies within the interval [1,1][-1, 1]. Therefore, the function is not continuous on the entire interval.
  3. Use of Theorem: Since the function is not continuous on the interval [1,1][-1, 1] due to the discontinuity at x=0x = 0, we cannot use the intermediate value theorem to guarantee that there is a solution to the equation h(x)=0.5h(x) = 0.5 on this interval.

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