Question

True or False The average value of a function \(f\) on \([ a , b ]\) always lies between \(f ( a )\) and \(f ( b ) .\) Justify your answer.

Short answer

False, the average value of a function on an interval does not necessarily lie between the function's values at the endpoints of the interval.

Step-by-step solution

Understanding the Mean Value Theorem

The Mean Value Theorem states that if a function \(f\) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there must be a point \(c\) in \((a, b)\) where the derivative of the function at that point is equal to the average rate of change of the function over the interval. This implies that for some \(c\) in \((a, b)\), we must have \(f(c) = \(\frac{1}{b-a}\) \(\int_{a}^{b} f(x) dx\).

Define conditions when the theorem fails

Note that the theorem fails when function \(f\) is not continuous and differentiable between points \(a\) and \(b\). For example, consider the function \(f(x) = |x|\) between the interval \([-1, 1]\). The function is not differentiable at \(x = 0\) as it has a corner at that point. Hence, the conditions of the theorem are not met and the theorem fails.

Conclusion

Hence, it is clear that the average value of a function on \([a, b]\) does not always lie between \(f(a)\) and \(f(b)\). This can be further emphasized by particular functions where the average value lies outside the function's values at the endpoints of the interval.