Problem 34
Question
Explain why the Mean Value Theorem does not apply to the function \(f\) on the interval \([0,6] .\) \(f(x)=|x-3|\)
Step-by-Step Solution
Verified Answer
The Mean Value Theorem isn't applicable to the function \(f(x)= |x-3|\) over the interval \([0,6]\) because this function isn't differentiable at \(x=3\), which lies inside this interval.
1Step 1: Understand the Function
The given function is an absolute value function which is written as \(f(x)= |x-3|\). This function forms a 'V' shape on the interval \([0,6]\) with a vertex at the point \((3,0)\).
2Step 2: Check Differentiability of Function
The function is differentiable everywhere except at point \(x=3\), because at \(x=3\) the function is not smooth, and it makes a 'sharp turn'. This breaches the conditions for applying the MVT, which requires the function to be differentiable over the entire interval.
3Step 3: Conclude the Result
Since the function \(f(x)= |x-3|\) is not differentiable at \(x=3\), which lies inside the interval \((0,6)\), hence the Mean Value Theorem cannot be applied to it on the interval \([0,6]\).
Key Concepts
Absolute Value FunctionDifferentiabilityCalculus Principles
Absolute Value Function
When talking about an absolute value function, such as the one presented in the exercise \(f(x) = |x-3|\), it is essential to grasp its graphical representation and behavior. This type of function is characterized by its distinctive 'V' shape on the Cartesian plane.
The function's formula computes the absolute value, which reflects the distance of a number from zero on a number line, disregarding its sign. The graph of \(f(x) = |x-3|\) pivots at the point where the expression inside the absolute value becomes zero, known in this case as the vertex \(\left(3,0\right)\).
For \(x\) values less than 3, the slope of the graph is negative, and for \(x\) values greater than 3, the slope is positive. This creates a sharp corner at the vertex, which is crucial when analyzing the function under calculus concepts such as differentiability and the Mean Value Theorem.
The function's formula computes the absolute value, which reflects the distance of a number from zero on a number line, disregarding its sign. The graph of \(f(x) = |x-3|\) pivots at the point where the expression inside the absolute value becomes zero, known in this case as the vertex \(\left(3,0\right)\).
For \(x\) values less than 3, the slope of the graph is negative, and for \(x\) values greater than 3, the slope is positive. This creates a sharp corner at the vertex, which is crucial when analyzing the function under calculus concepts such as differentiability and the Mean Value Theorem.
Differentiability
Differentiability is a cornerstone concept in calculus that pertains to the smoothness of a function's graph. For a function to be differentiable at a certain point, it must have a derivative at that point, meaning that there must be a single, well-defined tangent.
Remember, differentiability implies continuity, but the converse isn't always true. Therefore, while the function \(f(x) = |x-3|\) is continuous over the entire interval \[0,6\], it fails to be differentiable at the critical point \(x=3\). This nuance plays a pivotal role when applying theorems from calculus.
Sharp Turns and Differentiability
In the context of our absolute value function \(f(x) = |x-3|\), the sharp turn at \(x=3\) indicates the absence of such a tangent. Instead, the function has a cusp at \(x=3\), leading to an abrupt change in direction. This is what breaks the differentiability of the function there.Remember, differentiability implies continuity, but the converse isn't always true. Therefore, while the function \(f(x) = |x-3|\) is continuous over the entire interval \[0,6\], it fails to be differentiable at the critical point \(x=3\). This nuance plays a pivotal role when applying theorems from calculus.
Calculus Principles
Calculus principles, including the Mean Value Theorem (MVT), shape our understanding of function behavior over an interval. The Mean Value Theorem, in particular, provides a link between derivatives and increments of functions over intervals.
Applying these calculus principles to the given function \(f(x) = |x-3|\) over the interval \[0,6\], we hit a roadblock due to the lack of differentiability at the point \(x=3\). Even though the function meets the continuity requirement of the MVT, its failure to be differentiable throughout the entire interval prevents us from drawing conclusions that the MVT would normally allow.
It's through intricacies like this that calculus principles reveal the deep interconnections between the behavior of functions and the geometric and analytic properties they possess.
The Essence of the Mean Value Theorem
The MVT states that if a function \(f\) is continuous on a closed interval \[a,b\] and differentiable on the open interval \(a,b\), then there exists at least one point \(c\) in \(a,b\) where the instantaneous rate of change (derivative) is equal to the average rate of change over \[a,b\].Applying these calculus principles to the given function \(f(x) = |x-3|\) over the interval \[0,6\], we hit a roadblock due to the lack of differentiability at the point \(x=3\). Even though the function meets the continuity requirement of the MVT, its failure to be differentiable throughout the entire interval prevents us from drawing conclusions that the MVT would normally allow.
It's through intricacies like this that calculus principles reveal the deep interconnections between the behavior of functions and the geometric and analytic properties they possess.
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