Make H(x) Continuous: Find Values For 'a' And 'b'

Make h(x) Continuous: Find Values for 'a' and 'b'

Continuity is a fundamental concept in calculus, ensuring that a function's graph can be drawn without lifting your pen. For a function to be continuous at a point, three conditions must be met: the function must be defined at that point, the limit of the function must exist at that point, and the value of the function at that point must equal the limit. In this article, we'll explore how to ensure a piecewise function, $h(x)$, is continuous over its entire domain by determining the correct values for the constants $a$ and $b$. Understanding continuity is crucial for further mathematical explorations, including differentiation and integration, as these operations rely on the smooth, unbroken nature of functions. When we encounter functions defined in pieces, like the $h(x)$ we'll be working with, ensuring continuity often involves solving for unknown parameters at the points where the function definition changes. This process helps us bridge the gaps and create a single, coherent function from disparate parts. We'll break down the conditions for continuity and apply them systematically to our specific problem, making the abstract concept of continuity tangible and solvable.

Understanding Continuity Conditions

Let's first solidify our understanding of what it means for a function $f(x)$ to be continuous at a point $c$. There are three key conditions that must all be satisfied:

1. $f(c)$ is defined: This means that when you plug $c$ into the function, you get a real number output. There are no holes or undefined points at $c$ itself.
2. oldsymbol{ ext{lim}_{x o c}} f(x) exists: For the limit to exist, the function must approach the same value from both the left side (as $x$ approaches $c$ from values less than $c$) and the right side (as $x$ approaches $c$ from values greater than $c$). In mathematical terms, this means oldsymbol{ ext{lim}_{x o c^-}} f(x) = oldsymbol{ ext{lim}_{x o c^+}} f(x).
3. oldsymbol{ ext{lim}_{x o c}} f(x) = f(c): This is the crucial linking condition. The value the function approaches as $x$ gets closer and closer to $c$ must be exactly the same as the actual value of the function at $c$. If any of these conditions fail, the function is discontinuous at $c$. We often categorize discontinuities as jump discontinuities, removable discontinuities (holes), or infinite discontinuities (asymptotes).

For our piecewise function $h(x)$, we are given:

$h(x)=\\begin{cases}\ x^3, & x<0 \ a, & x=0 \ \sqrt{x}, & 0 < x < 4 \ b, & x=4 \ 4-\frac{1}{2} x, & x>4 \\ ext{end{cases}

We need to ensure $h(x)$ is continuous for all values of $x$. The points where we need to pay special attention are where the definition of the function changes: at $x=0$ and at $x=4$. These are the potential locations for discontinuities.

Ensuring Continuity at $x=0$

To make $h(x)$ continuous at $x=0$, we must satisfy the three conditions mentioned above. Let's apply them:

1. $h(0)$ is defined: From the definition of $h(x)$, we see that $h(0) = a$. So, as long as $a$ is a real number, this condition is met.

2. oldsymbol{ ext{lim}_{x o 0}} h(x) exists: For the limit to exist at $x=0$, the limit from the left must equal the limit from the right.

   • Limit from the left (oldsymbol{ ext{lim}_{x o 0^-}} h(x)): As $x$ approaches 0 from values less than 0, we use the definition $h(x) = x^3$. So, oldsymbol{ ext{lim}_{x o 0^-}} h(x) = oldsymbol{ ext{lim}_{x o 0^-}} x^3 = 0^3 = 0.
   • Limit from the right (oldsymbol{ ext{lim}_{x o 0^+}} h(x)): As $x$ approaches 0 from values greater than 0, we use the definition $h(x) = \sqrt{x}$. So, oldsymbol{ ext{lim}_{x o 0^+}} h(x) = oldsymbol{ ext{lim}_{x o 0^+}} \sqrt{x} = \sqrt{0} = 0. Since the limit from the left (0) equals the limit from the right (0), the limit oldsymbol{ ext{lim}_{x o 0}} h(x) exists and is equal to 0.

3. oldsymbol{ ext{lim}_{x o 0}} h(x) = h(0): For continuity, the limit we found (0) must equal the function's value at $x=0$. We know $h(0) = a$. Therefore, we must have $0 = a$.

So, for $h(x)$ to be continuous at $x=0$, we must set $a=0$. This ensures that the piece $x^3$ smoothly connects to the defined point $a$ at $x=0$, and the piece $\sqrt{x}$ also starts from that same point.

Ensuring Continuity at $x=4$

Now, let's apply the same continuity conditions to the point $x=4$.

1. $h(4)$ is defined: From the definition of $h(x)$, we see that $h(4) = b$. So, as long as $b$ is a real number, this condition is met.

2. oldsymbol{ ext{lim}_{x o 4}} h(x) exists: Again, for the limit to exist at $x=4$, the limit from the left must equal the limit from the right.

   • Limit from the left (oldsymbol{ ext{lim}_{x o 4^-}} h(x)): As $x$ approaches 4 from values less than 4, we use the definition $h(x) = \sqrt{x}$ (since the domain for $\sqrt{x}$ is given as $0 < x < 4$). So, oldsymbol{ ext{lim}_{x o 4^-}} h(x) = oldsymbol{ ext{lim}_{x o 4^-}} \sqrt{x} = \sqrt{4} = 2.
   • Limit from the right (oldsymbol{ ext{lim}_{x o 4^+}} h(x)): As $x$ approaches 4 from values greater than 4, we use the definition $h(x) = 4 - \frac{1}{2} x$. So, oldsymbol{ ext{lim}_{x o 4^+}} h(x) = oldsymbol{ ext{lim}_{x o 4^+}} (4 - \frac{1}{2} x) = 4 - \frac{1}{2}(4) = 4 - 2 = 2. Since the limit from the left (2) equals the limit from the right (2), the limit oldsymbol{ ext{lim}_{x o 4}} h(x) exists and is equal to 2.

3. oldsymbol{ ext{lim}_{x o 4}} h(x) = h(4): For continuity at $x=4$, the limit we found (2) must equal the function's value at $x=4$. We know $h(4) = b$. Therefore, we must have $2 = b$.

So, for $h(x)$ to be continuous at $x=4$, we must set $b=2$. This ensures that the piece $\sqrt{x}$ smoothly connects to the defined point $b$ at $x=4$, and the piece $4 - \frac{1}{2} x$ also starts from that same point.

The Complete Continuous Function

By determining the values of $a$ and $b$, we have successfully made the function $h(x)$ continuous over its entire domain. The completed function is:

$h(x)=\\begin{cases}\ x^3, & x<0 \ 0, & x=0 \ \sqrt{x}, & 0 < x < 4 \ 2, & x=4 \ 4-\frac{1}{2} x, & x>4 \\end{cases}

This means that if you were to graph this function, you could draw it without lifting your pen. The pieces of the function meet at precisely the right points ($x=0$ and $x=4$) to create a single, unbroken curve.

Conclusion

We've seen how the fundamental conditions of continuity—that the function is defined, the limit exists, and the function's value equals its limit at a point—allow us to solve for unknown constants in a piecewise function. By carefully evaluating the left-hand and right-hand limits at the points where the function definition changes ($x=0$ and $x=4$), we found that $a=0$ and $b=2$ are necessary to bridge the pieces of $h(x)$ and ensure overall continuity. This process is a cornerstone in understanding function behavior and is essential for more advanced calculus topics. Mastering continuity problems like this one builds a strong foundation for exploring the intricacies of mathematical functions.

For further exploration into the fascinating world of calculus and continuous functions, you can visit:

• Khan Academy's comprehensive section on limits and continuity, offering detailed explanations and practice problems.
• Paul's Online Math Notes provides in-depth notes and examples on calculus topics, including continuity.