Continuity And Differentiability Of Composite Functions Analysis Of F(g(x))

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In the fascinating realm of mathematical analysis, understanding the behavior of functions is paramount. This article delves into the intricate world of composite functions, specifically focusing on the function f(g(x)), where f(x) and g(x) are piecewise-defined functions. We will explore the conditions for continuity and differentiability of such composite functions, providing a comprehensive analysis and insights. Our exploration is anchored on the following definitions:

  • f(x) = egin{cases} x+2, & 0 < x < 2 \ 6-x, & x "> 2

Let's first unpack the functions f(x) and g(x). The function f(x) is defined in two pieces. For x values between 0 and 2 (exclusive), it behaves like the linear function x + 2. However, when x is greater than or equal to 2, it transitions to another linear function, 6 - x. This piecewise nature is crucial in understanding the overall behavior of f(x).

Now, let's turn our attention to g(x). This function also has a piecewise definition, but it's defined in terms of trigonometric functions. For x values between 0 (inclusive) and π/4 (exclusive), g(x) takes the form 1 + tan(x). On the other hand, for x values between π/4 (inclusive) and π (exclusive), it follows the rule 3 - cot(x). The presence of trigonometric functions introduces additional complexity, as we need to consider their behavior within the specified intervals.

The core challenge lies in understanding how these two functions interact when composed. The composite function f(g(x)) means we're plugging the output of g(x) into f(x). This creates a chain reaction, where the behavior of g(x) directly influences the input to f(x), and consequently, the final output. To analyze f(g(x)), we need to carefully consider the range of g(x) and how it aligns with the domain of f(x). This is particularly important at the points where the pieces of f(x) and g(x) change, as these are potential points of discontinuity or non-differentiability.

Dissecting the Continuity of f(g(x))

To rigorously determine the continuity of f(g(x)), we need to examine its behavior across its entire domain. This involves breaking down the analysis into intervals determined by the piecewise definitions of f(x) and g(x). Continuity at a point requires three conditions to be met: the function must be defined at the point, the limit of the function must exist at the point, and the limit must equal the function's value at the point. Any failure in these conditions leads to a discontinuity.

The critical points to consider are where the definitions of f(x) and g(x) change. For f(x), this point is x = 2. For g(x), the key point is x = π/4. These points act as boundaries, dividing the domain into subintervals where the behavior of the composite function might differ significantly. We need to meticulously analyze f(g(x)) in each of these subintervals.

Let's consider the interval 0 ≤ x < π/4. In this interval, g(x) = 1 + tan(x). As x varies from 0 to π/4, tan(x) increases from 0 to 1, so g(x) ranges from 1 to 2. Now, we need to see how this range interacts with f(x). When g(x) is between 1 and 2, we use the first piece of f(x), which is f(x) = x + 2. Therefore, in this interval, f(g(x)) = g(x) + 2 = (1 + tan(x)) + 2 = 3 + tan(x). This part of f(g(x)) is continuous because both the tangent function and the constant function are continuous in this interval.

Next, we consider the interval π/4 ≤ x < π. Here, g(x) = 3 - cot(x). As x varies from π/4 to π, cot(x) decreases from 1 to negative infinity. Consequently, g(x) increases from 2 to positive infinity. This is where things get interesting, because g(x) now enters the domain where the second piece of f(x) applies, which is f(x) = 6 - x. So, for these values of x, f(g(x)) = 6 - g(x) = 6 - (3 - cot(x)) = 3 + cot(x). Again, this part of f(g(x)) is continuous in its interval because the cotangent function is continuous in this interval.

However, the critical point for continuity analysis is at x = π/4, where the definition of g(x) changes. We need to ensure that the left-hand limit and the right-hand limit of f(g(x)) at x = π/4 exist and are equal. The left-hand limit is lim(x→π/4-) f(g(x)) = lim(x→π/4-) (3 + tan(x)) = 3 + tan(π/4) = 3 + 1 = 4. The right-hand limit is lim(x→π/4+) f(g(x)) = lim(x→π/4+) (3 + cot(x)) = 3 + cot(π/4) = 3 + 1 = 4. Since the left-hand limit equals the right-hand limit, and both are equal to the value of f(g(π/4)), the composite function is continuous at x = π/4.

To solidify our understanding, it's essential to consider the implications of continuity. A continuous function allows for smooth transitions, meaning that small changes in the input result in small changes in the output. This property is crucial in many applications, from modeling physical phenomena to designing stable control systems.

Evaluating the Differentiability of f(g(x))

While continuity is a fundamental property, differentiability takes our analysis a step further. Differentiability implies that a function has a derivative at a particular point, which represents the instantaneous rate of change of the function at that point. Geometrically, the derivative is the slope of the tangent line to the function's graph. A function must be continuous at a point to be differentiable at that point, but the converse is not always true.

To investigate the differentiability of f(g(x)), we need to compute the derivative of the function in each interval where it is defined and then check if the derivatives match at the critical points. The chain rule is essential here, as it provides a way to differentiate composite functions. The chain rule states that the derivative of f(g(x)) with respect to x is f'(g(x)) * g'(x).

In the interval 0 < x < π/4, we found that f(g(x)) = 3 + tan(x). The derivative of this function is sec²(x). This derivative exists and is continuous in this interval.

In the interval π/4 < x < π, we have f(g(x)) = 3 + cot(x). The derivative here is -csc²(x), which also exists and is continuous in this interval.

The critical point for differentiability is again x = π/4. We need to check if the left-hand derivative and the right-hand derivative at x = π/4 are equal. The left-hand derivative is lim(x→π/4-) sec²(x) = sec²(π/4) = (√2)² = 2. The right-hand derivative is lim(x→π/4+) -csc²(x) = -csc²(π/4) = -(√2)² = -2. Since the left-hand derivative (2) is not equal to the right-hand derivative (-2), the function f(g(x)) is not differentiable at x = π/4.

The non-differentiability at x = π/4 implies that the graph of f(g(x)) has a sharp corner or a cusp at this point. This means that the function changes direction abruptly, and a unique tangent line cannot be defined at this point. This kind of behavior is characteristic of functions that are pieced together from different components with differing slopes.

Conclusion: A Comprehensive Analysis

In conclusion, our in-depth analysis of the composite function f(g(x)) reveals that while it is continuous on the interval [0, π], it is not differentiable at x = π/4. This outcome underscores the subtle interplay between continuity and differentiability. A function can be continuous without being differentiable, as demonstrated by the sharp corner in the graph of f(g(x)) at x = π/4.

This exploration provides valuable insights into the behavior of composite functions and reinforces the importance of careful analysis when dealing with piecewise-defined functions. Understanding these concepts is essential for tackling more advanced topics in calculus and analysis. The principles discussed here extend to a wide range of applications, from physics and engineering to economics and computer science, where the modeling of complex systems often involves the composition of functions.

Let f(x) = egin{cases} x+2, & 0 < x < 2 \ 6-x, & x "> 2

Analysis of Continuity of f(g(x))

To determine if f(g(x)) is continuous on the interval [0, π], we need to examine the continuity of f(x) and g(x) individually, and then analyze the behavior of their composition. A composite function f(g(x)) is continuous at a point c if g is continuous at c and f is continuous at g(c). Let’s break this down step-by-step.

Continuity of g(x)

First, let's analyze the continuity of g(x). The function g(x) is defined as follows:

  • g(x) = 1 + tan(x), for 0 ≤ x < π/4
  • g(x) = 3 - cot(x), for π/4 ≤ x < π

Each piece of g(x) involves trigonometric functions, specifically tan(x) and cot(x). We know that tan(x) is continuous on the interval [0, π/4) and cot(x) is continuous on the interval (π/4, π). The potential issue is at x = π/4, where the definition of g(x) changes. To ensure continuity at this point, the left-hand limit and the right-hand limit must be equal, and also equal to the function value at that point.

Let’s compute the left-hand limit:

lim (x→π/4-) g(x) = lim (x→π/4-) (1 + tan(x)) = 1 + tan(π/4) = 1 + 1 = 2

Now, the right-hand limit:

lim (x→π/4+) g(x) = lim (x→π/4+) (3 - cot(x)) = 3 - cot(π/4) = 3 - 1 = 2

And the value of g(x) at x = π/4: g(π/4) = 3 - cot(π/4) = 3 - 1 = 2

Since the left-hand limit, right-hand limit, and the function value at x = π/4 are all equal to 2, g(x) is continuous at x = π/4. Therefore, g(x) is continuous on the interval [0, π).

Continuity of f(x)

Next, we assess the continuity of f(x), which is defined as:

  • f(x) = x + 2, for 0 < x < 2
  • f(x) = 6 - x, for x ≥ 2

Both pieces of f(x) are linear functions and are continuous within their respective intervals. We need to check the continuity at x = 2.

The left-hand limit is:

lim (x→2-) f(x) = lim (x→2-) (x + 2) = 2 + 2 = 4

The right-hand limit is:

lim (x→2+) f(x) = lim (x→2+) (6 - x) = 6 - 2 = 4

And the function value at x = 2: f(2) = 6 - 2 = 4

Since the left-hand limit, right-hand limit, and the function value at x = 2 are all equal to 4, f(x) is continuous at x = 2. Thus, f(x) is continuous for x > 0.

Continuity of f(g(x))

Now, we analyze the continuity of the composite function f(g(x)). We know that g(x) is continuous on [0, π) and f(x) is continuous for x > 0. The range of g(x) needs to be considered for the domain of f(x).

For 0 ≤ x < π/4, g(x) = 1 + tan(x). As x varies from 0 to π/4, g(x) varies from 1 to 2. Since f(x) is continuous on this range, f(g(x)) is continuous on [0, π/4).

For π/4 ≤ x < π, g(x) = 3 - cot(x). As x varies from π/4 to π, g(x) varies from 2 to positive infinity. Since f(x) is continuous for x ≥ 2, f(g(x)) is continuous on [π/4, π).

The point of concern is x = π/4. We already know that g(π/4) = 2. We also know that f(2) = 4. Now we need to check the left-hand and right-hand limits of f(g(x)) as x approaches π/4.

For the left-hand limit:

lim (x→π/4-) f(g(x)) = lim (x→π/4-) f(1 + tan(x)) = lim (x→π/4-) ((1 + tan(x)) + 2) = 1 + 1 + 2 = 4

For the right-hand limit:

lim (x→π/4+) f(g(x)) = lim (x→π/4+) f(3 - cot(x)) = lim (x→π/4+) (6 - (3 - cot(x))) = 6 - (3 - 1) = 4

Since the left-hand limit, right-hand limit, and f(g(π/4)) are all equal to 4, f(g(x)) is continuous at x = π/4. Thus, f(g(x)) is continuous on [0, π).

Analysis of Differentiability of f(g(x))

To examine the differentiability of f(g(x)), we need to find the derivative of f(g(x)) and check for any points where the derivative might not exist. This typically occurs at points where the functions are pieced together, or where the individual functions are not differentiable.

Differentiability Using the Chain Rule

We'll use the chain rule, which states that if we have a composite function f(g(x)), its derivative is given by:

(f(g(x)))' = f'(g(x)) * g'(x)

First, we find the derivatives of f(x) and g(x) individually.

Derivatives of f(x)

f(x) is defined piecewise, so we differentiate each piece:

  • If 0 < x < 2, f(x) = x + 2, so f'(x) = 1
  • If x > 2, f(x) = 6 - x, so f'(x) = -1

The derivative f'(x) is also piecewise:

f'(x) = egin{cases} 1, & 0 < x < 2 \ -1, & x "> 2

Note that f'(x) does not exist at x = 2.

Derivatives of g(x)

Similarly, we find the derivative of g(x):

  • If 0 ≤ x < π/4, g(x) = 1 + tan(x), so g'(x) = sec²(x)
  • If π/4 < x < π, g(x) = 3 - cot(x), so g'(x) = csc²(x)

Thus, g'(x) is given by:

g'(x) = egin{cases} sec²(x), & 0 "> π/4

Note that g'(x) does not exist at x = π/4.

Derivative of f(g(x))

Now we find the derivative of f(g(x)) using the chain rule. We need to consider the intervals and match the pieces appropriately.

  1. For 0 < x < π/4:
    • g(x) = 1 + tan(x), which lies in the interval (1, 2) for f(x), so f'(g(x)) = 1.
    • g'(x) = sec²(x)
    • Therefore, (f(g(x)))' = f'(g(x)) * g'(x) = 1 * sec²(x) = sec²(x).
  2. For π/4 < x < π:
    • g(x) = 3 - cot(x), which lies in the interval (2, ∞) for f(x), so f'(g(x)) = -1.
    • g'(x) = csc²(x)
    • Therefore, (f(g(x)))' = f'(g(x)) * g'(x) = -1 * csc²(x) = -csc²(x).

So, the derivative of f(g(x)) is:

(f(g(x)))' = egin{cases} sec²(x), & 0 "> π/4

Checking Differentiability at x = π/4

To determine if f(g(x)) is differentiable at x = π/4, we check if the left-hand derivative and the right-hand derivative are equal.

The left-hand derivative at x = π/4:

lim (x→π/4-) (f(g(x)))' = lim (x→π/4-) sec²(x) = sec²(π/4) = (√2)² = 2

The right-hand derivative at x = π/4:

lim (x→π/4+) (f(g(x)))' = lim (x→π/4+) (-csc²(x)) = -csc²(π/4) = -(√2)² = -2

Since the left-hand derivative (2) is not equal to the right-hand derivative (-2), f(g(x)) is not differentiable at x = π/4.

Conclusion on Differentiability

In summary, f(g(x)) is differentiable on the intervals (0, π/4) and (π/4, π), but it is not differentiable at x = π/4.

  • f(g(x)) is continuous on [0, π].
  • f(g(x)) is not differentiable at x = π/4.

This analysis provides a comprehensive understanding of the continuity and differentiability of the composite function f(g(x)).