Problem 18
Question
Choose \(A\) and \(B\) so as to make the function \(f(x)\) continuous at \(x=\pm
\frac{\pi}{2}\)
$$
\begin{aligned}
f(x) &=-2 \sin x, \quad x \leq-\frac{\pi}{2} \\
&=A \sin x+B, \quad-\frac{\pi}{2}
Step-by-Step Solution
Verified Answer
To make the function \(f(x)\) continuous at \(x = \pm \frac{\pi}{2}\), we need to find values of A and B that satisfy the conditions of left and right limits being equal at these points. By solving the system of equations for these conditions, we find that \(A = -2\) and \(B = 0\). Therefore, the function is continuous at \(x = \pm \frac{\pi}{2}\) when written as:
$$
\begin{aligned}
f(x) &= -2 \sin x, \quad x \leq -\frac{\pi}{2} \\
&= -2 \sin x, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}
\end{aligned}
$$
1Step 1: Setup Limits
Set up the limits for both sides of the function at both points.
Left limit at \(x = -\frac{\pi}{2}\):
\( \underset{x \to -\frac{\pi}{2}^{-}}{lim} -2 \sin x \)
Right limit at \(x = -\frac{\pi}{2}\):
\( \underset{x \to -\frac{\pi}{2}^{+}}{lim} (A \sin x + B) \)
Left limit at \(x = \frac{\pi}{2}\):
\( \underset{x \to \frac{\pi}{2}^{-}}{lim} (A \sin x + B) \)
At these points, the left and right limits must be equal for the function to be continuous.
2Step 2: Equate the left and right limits at \(x = -\frac{\pi}{2}\)
Equate the left and right limits at \(x = -\frac{\pi}{2}\):
\(-2 \sin (-\frac{\pi}{2}) = A \sin (-\frac{\pi}{2}) + B\)
Now, we know that \(\sin (-\frac{\pi}{2}) = -1\), so:
\(2 = - A + B\)
This is our first equation.
3Step 3: Equate the left and right limits at \(x = \frac{\pi}{2}\)
Equate the left and right limits at \(x = \frac{\pi}{2}\):
\(A \sin (\frac{\pi}{2}) + B = A \sin (\frac{\pi}{2})\)
Since \(\sin (\frac{\pi}{2}) = 1\), we get:
\(A + B = A\)
This is our second equation.
4Step 4: Solve the system of equations
We now have a system of two equations with two variables, A and B:
1) \(2 = - A + B\)
2) \(A + B = A\)
Solving the second equation for B gives us:
\(B = 0\)
Now substitute B back into equation 1:
\(2 = - A + 0\)
Solve for A:
\(A = -2\)
5Step 5: Final Answer
Now we have the values for A and B that will make the function continuous at \(x = \pm \frac{\pi}{2}\):
\(A = -2\), and \(B = 0\)
So the function can be written as:
$$
\begin{aligned}
f(x) &= -2 \sin x, \quad x \leq -\frac{\pi}{2} \\
&= -2 \sin x, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}
\end{aligned}
$$
Key Concepts
Limit of a FunctionContinuous FunctionSolving System of Equations
Limit of a Function
Understanding the limit of a function is like peering as close as possible to a particular point on a graph without actually touching it. This concept is critical in calculus because it helps us handle values that are approaching some number, but do not necessarily ever reach it.
For instance, consider the equation \(\lim_{{x \to c}} f(x)\) which implies we are interested to see what value the function \(f(x)\) is nearing as \(x\) approaches some number \(c\). In the exercise, we looked at the behavior of two pieces of a piecewise function as they approach \(\pm\frac{\pi}{2}\) from both sides.
Why does this matter? Well, by finding these limits, we can determine key behavior of the function which is essential for ensuring continuity at those points. It's the mathematical equivalent of stitching together two fabrics seamlessly. If the limits on both sides of a point match up, the function can pass the 'continuous' test at that point.
For instance, consider the equation \(\lim_{{x \to c}} f(x)\) which implies we are interested to see what value the function \(f(x)\) is nearing as \(x\) approaches some number \(c\). In the exercise, we looked at the behavior of two pieces of a piecewise function as they approach \(\pm\frac{\pi}{2}\) from both sides.
Why does this matter? Well, by finding these limits, we can determine key behavior of the function which is essential for ensuring continuity at those points. It's the mathematical equivalent of stitching together two fabrics seamlessly. If the limits on both sides of a point match up, the function can pass the 'continuous' test at that point.
Continuous Function
A function is continuous at a point when you can draw it without lifting your pen off the paper. In other words, there are no holes, jumps, or breaks at that point. Mathematically, a function \( f(x) \) is continuous at \( x = c \) if the following three conditions are met:
1. The function is defined at \( c \) (\( f(c) \) exists).2. The limit of \( f(x) \) as \( x \) approaches \( c \) exists.3. The limit of \( f(x) \) as \( x \) approaches \( c \) equals \( f(c) \) (\( \lim_{{x \to c}} f(x) = f(c) \) ).
In our exercise, we wanted to make sure the function was continuous at \( \pm\frac{\pi}{2} \). Therefore, we calculated the limits from both sides of these points and set about finding values for \( A \) and \( B \) so that the function satisfied the definition of continuity. Thus, making the function glide smoothly at these junctures was much like finding the perfect alignment so that there's no disruption as one traverses the curve on the graph.
1. The function is defined at \( c \) (\( f(c) \) exists).2. The limit of \( f(x) \) as \( x \) approaches \( c \) exists.3. The limit of \( f(x) \) as \( x \) approaches \( c \) equals \( f(c) \) (\( \lim_{{x \to c}} f(x) = f(c) \) ).
In our exercise, we wanted to make sure the function was continuous at \( \pm\frac{\pi}{2} \). Therefore, we calculated the limits from both sides of these points and set about finding values for \( A \) and \( B \) so that the function satisfied the definition of continuity. Thus, making the function glide smoothly at these junctures was much like finding the perfect alignment so that there's no disruption as one traverses the curve on the graph.
Solving System of Equations
Sometimes in mathematics, we encounter multiple unknowns and need a strategy to decipher their values. This is where solving a system of equations comes into play. It's like being a detective with a number of clues and trying to solve the mystery—each equation acts as a crucial clue.
In our specific puzzle, we had two equations which arose from setting the left and right limits equal at both \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), as the function needs to be continuous at these points. The equations formed a system:
1. \( 2 = - A + B \)
2. \( A + B = A \)
We used substitution and elimination methods to find the values of \( A \) and \( B \). First, we figured out \( B \) by observing that the second equation implied \( B \) must be zero, and then we plugged it into the first equation to solve for \( A \). These values ensured our function's continuity at the specified points and gave us a clearer picture of the graph's overall landscape.
In our specific puzzle, we had two equations which arose from setting the left and right limits equal at both \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), as the function needs to be continuous at these points. The equations formed a system:
1. \( 2 = - A + B \)
2. \( A + B = A \)
We used substitution and elimination methods to find the values of \( A \) and \( B \). First, we figured out \( B \) by observing that the second equation implied \( B \) must be zero, and then we plugged it into the first equation to solve for \( A \). These values ensured our function's continuity at the specified points and gave us a clearer picture of the graph's overall landscape.
Other exercises in this chapter
Problem 16
\(\begin{aligned} f(x) &=\frac{1-\cos 4 x}{x^{2}}, x0 . \end{aligned}\)
View solution Problem 17
For what value of \(a\), the function \(\begin{aligned} f(x) &=x^{a} \sin \frac{1}{x}, \quad x \neq 0 \\ &=0, \quad x=0 \end{aligned}\)
View solution Problem 19
Let \(\begin{aligned} f(x) &=(1+|\sin x|) \sqrt{\sin x}, \quad-\frac{\pi}{6}
View solution Problem 21
Determine \(a, b\) and \(c\) for which the function \(f(x)=\frac{\sin (a+1) x+\sin x}{x}, x0\)
View solution