Question

Let \(f(x)=\frac{|x|}{x} .\) Then \(f(-2)=-1\) and \(f(2)=1 .\) Therefore \(f(-2)<0

Short answer

Explain your answer. Answer: No, the fact that there is no value c for which \(f(c)=0\) does not violate the Intermediate Value Theorem. This is because the function \(f(x)=\frac{|x|}{x}\) is not continuous on the interval from -2 to 2 due to its undefined value at x=0. The Intermediate Value Theorem can only be applied if the function is continuous on the considered interval.

Step-by-step solution

Examine the function and its continuity

Let's examine the given function \(f(x)=\frac{|x|}{x}\). We know that the absolute value is continuous everywhere; however, the function is undefined when x=0 because division by 0 is not allowed. Thus, the given function is not continuous on the interval from -2 to 2.

Consider the Intermediate Value Theorem

The Intermediate Value Theorem states that if a function is continuous on an interval [a, b], and a value k lies between the function's values at a and b (i.e., \(f(a)k>f(b)\)), then there exists a point 'c' in the interval where the function takes that intermediate value (i.e., \(f(c)=k\)). In this problem, we have \(f(-2)=-1\), \(f(2)=1\), and we're looking for a value \(c\) between -2 and 2 where \(f(c)=0\). However, since the function is not continuous on the interval, we cannot directly apply the Intermediate Value Theorem.

Conclusion

The fact that there is no value \(c\) between -2 and 2 for which \(f(c)=0\) does not violate the Intermediate Value Theorem because the function \(f(x)=\frac{|x|}{x}\) is not continuous on the interval from -2 to 2. The Intermediate Value Theorem can only be applied if the function is continuous on the considered interval. So in this case, the Intermediate Value Theorem does not apply.