Let f(x) = {x+a if x

Q: Wait, why do we check $x = -a$ for continuity? I thought we only check the transition points of the original functions.

That is a fantastic question! While $x=0$ is a transition point for both $f(x)$ and $g(x)$, the composite function $g(f(x))$ inherits new transition points. Because $g(x)$ changes its rule at $0$, the composite function $g(f(x))$ will change its rule whenever $f(x)$ crosses $0$. Since $f(x) = x+a$ for $x < 0$, the condition $f(x) = 0$ occurs at $x = -a$. Thus, $x = -a$ is a critical transition point for the composite function that we absolutely must check!

Q: Why does $f(x) = |x-1|$ mean we don't need to split the domain for $x \geq 0$?

Great observation. The definition of $g(y)$ changes at $y=0$. For $x \geq 0$, $f(x) = |x-1|$. By definition, the absolute value of any real number is always non-negative, meaning $f(x) \geq 0$ for all $x \geq 0$. Since the input to $g$ is always non-negative in this region, we only ever need to use the second branch of $g(y)$, which is $(y-1)^2 + b$. No splitting is required because the input never crosses the $0$ threshold!

Q: Is it always true that if a function is continuous, it's differentiable? Why did we have to check differentiability separately?

Not at all! Continuity is a necessary condition for differentiability, but it is not sufficient. A function can be continuous but have a 'sharp corner' or 'kink' where the derivative is undefined (like the absolute value function at $x=0$). We checked differentiability separately by ensuring the left-hand derivative and right-hand derivative at $x=0$ were equal. If they hadn't matched, the function would be continuous but not differentiable.

Question

Let $f(x) = \begin{cases} x+a & 	ext{if } x < 0 \ |x-1| & 	ext{if } x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x+1 & 	ext{if } x < 0 \ (x-1)^2+b & 	ext{if } x \geq 0 \end{cases}$ where $a$ and $b$ are non-negative real numbers. Determine the composite function $g \circ f$. If $(g \circ f)(x)$ is continuous for all real $x$, determine the values of $a$ and $b$. Further, for these values of $a$ and $b$, is $g \circ f$ differentiable at $x=0$? Justify your answer.

Sigma Percentile · Accepted Answer

### The Relay Race of Functions

Imagine you are standing at the starting line of a relay race. In mathematics, a composite function $g(f(x))$ is exactly that—a relay race. The function $f(x)$ runs the first leg, and its output is handed off as the input to $g(x)$.

To master this problem, we must understand not just the functions themselves, but the hand-off point. The function $g(x)$ is a bit temperamental; it changes its behavior entirely depending on whether its input is negative or non-negative.

This means the composite function $g(f(x))$ will change its behavior whenever $f(x)$ crosses the threshold of zero. Let us peel back the layers of this problem together.

### Navigating the Negative Domain

We begin by looking at the region where $x < 0$. In this territory, $f(x) = x+a$. But remember, $g(x)$ changes its rule at $0$.

So, we must ask: when is $f(x) < 0$ and when is $f(x) \geq 0$? If $x+a < 0$, which simplifies to $x < -a$, we must use the first branch of $g(y)$, which is $y+1$.

Substituting $f(x)$ for $y$, we get:
$$g(f(x)) = (x+a) + 1$$

However, if $x+a \geq 0$, which means $-a \leq x < 0$, we must switch to the second branch of $g(y)$, which is $(y-1)^2 + b$. Substituting $f(x)$ here, we get:
$$g(f(x)) = (x+a-1)^2 + b$$

We have successfully mapped the behavior of our composite function for all $x < 0$. It is a piecewise function within a piecewise function—a beautiful, nested structure.

### The Absolute Value Simplification

Now, let us shift our gaze to the right side of the y-axis, where $x \geq 0$. Here, $f(x) = |x-1|$.

A wonderful property of the absolute value function is that it is always non-negative. Since $f(x) \geq 0$ for all $x \geq 0$, we never have to worry about the first branch of $g(y)$.

We can confidently use the second branch, $(y-1)^2 + b$, for the entire region. Substituting $f(x) = |x-1|$, our composite function becomes:
$$g(f(x)) = (|x-1|-1)^2 + b$$

The complexity melts away, leaving us with a clear definition for the entire real line.

### The Continuity Puzzle

For $g(f(x))$ to be continuous everywhere, it must be 'glued' together perfectly at its transition points. We have two critical points to check: $x = -a$ and $x = 0$.

At $x = -a$, we equate the limits from the left and right. The left-hand limit is:
$$\lim_{x 	o -a^-} (x+a+1) = 1$$

The right-hand limit is:
$$\lim_{x 	o -a^+} ((x+a-1)^2+b) = (-a+a-1)^2+b = 1+b$$

Setting these equal, we find $1 = 1+b$, which immediately gives us **$b = 0$**.

With $b=0$ in hand, we move to $x = 0$. The left-hand limit is:
$$\lim_{x 	o 0^-} ((x+a-1)^2+0) = (a-1)^2$$

The right-hand limit is:
$$\lim_{x 	o 0^+} ((|x-1|-1)^2+0) = (|0-1|-1)^2 = 0$$

Equating these, we get $(a-1)^2 = 0$, which tells us **$a = 1$**. We have solved the mystery of the constants!

### The Final Smoothness Test

Finally, we test for differentiability at $x=0$. With $a=1$ and $b=0$, our function $h(x) = g(f(x))$ simplifies beautifully.

For $x$ near $0$, the left-hand side becomes $(x+1-1)^2 = x^2$. For the right-hand side, since $x$ is small and positive, $|x-1| = 1-x$, so:
$$h(x) = (1-x-1)^2 = (-x)^2 = x^2$$

Since $h(x) = x^2$ on both sides of zero, the function is not just continuous; it is smooth. The derivative $h'(x) = 2x$ exists at $x=0$ and is equal to $0$.

The relay race is complete, and the transition is seamless. You have navigated the piecewise logic, enforced the continuity constraints, and verified the smoothness of the result. **Well done!**

Visualized Solution

Understanding the Composite Function $g (f (x))$

Analyzing $x < 0$

Splitting the Domain for $x < 0$

Analyzing $x \geq 0$

Defining $g (f (x))$ for $x \geq 0$

Continuity at $x = - a$

Evaluating Limits at $x = - a$

Solving for $b$

Continuity at $x = 0$

Evaluating Limits at $x = 0$

Solving for $a$

Differentiability at $x = 0$

Function near $x = 0$

Conclusion on Differentiability

The Sigma Insight: Relationship Between Continuity and Differentiability

The Relay Race of Functions

Navigating the Negative Domain

The Absolute Value Simplification

The Continuity Puzzle

The Final Smoothness Test

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Visualized Solution

Understanding the Composite Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f(x))

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Splitting the Domain for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(f(x)) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≥0

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−a

Evaluating Limits at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−a

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Evaluating Limits at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Function near .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0