Answer :
To solve the problem step-by-step, let's first understand the function [tex]\( f(x) \)[/tex] and graph it.
The function is given as follows:
[tex]\[ f(x) = \begin{cases} 1 - x^2 & \text{if } x < 1 \\ -2x - 3 & \text{if } x \geq 1 \end{cases} \][/tex]
### Step 1: Graph the function
To graph the function, we need to do the following:
1. Graph the piece [tex]\( 1 - x^2 \)[/tex] for [tex]\( x < 1 \)[/tex]:
This is a downward-opening parabola. When [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex], and it opens downward with its vertex at [tex]\( (0,1) \)[/tex].
2. Graph the piece [tex]\( -2x - 3 \)[/tex] for [tex]\( x \geq 1 \)[/tex]:
This is a linear function with a slope of -2 and a y-intercept of -3. For [tex]\( x = 1 \)[/tex], [tex]\( y = -2(1) - 3 = -5 \)[/tex].
### Step 2: Determine boundaries and transitions
To better visualize the function's continuity at [tex]\( x = 1 \)[/tex], let's calculate a few critical points:
For [tex]\( x < 1 \)[/tex]:
[tex]\[ \begin{aligned} f(0) &= 1 - 0^2 = 1 \\ f(0.5) &= 1 - (0.5)^2 = 1 - 0.25 = 0.75 \\ f(0.9) &= 1 - (0.9)^2 = 1 - 0.81 = 0.19 \end{aligned} \][/tex]
For continuity at [tex]\( x = 1 \)[/tex]:
[tex]\[ \begin{aligned} \lim_{x \to 1^-} f(x) &= f(1^-)= 1 - 1^2 = 0 \\ f(1) &= -2(1) - 3 = -5 \\ \lim_{x \to 1^+} f(x) = f(1^+) &= -2(1 + \epsilon) - 3= -2 - 3 = -5, \end{aligned} \][/tex]
where+ [tex]\(\epsilon\)[/tex] is a small number approaching 0.
### Step 3: Summary and Conclusion
So, let's evaluate:
- From the left-hand side as [tex]\( x \)[/tex] approaches 1 ([tex]\( x<1 \)[/tex]), the limit is:
[tex]\[ \lim_{x \to 1^-} (1 - x^2) = 0. \][/tex]
- From the right-hand side ([tex]\( x \ge 1 \)[/tex]), the limit is:
[tex]\[ \lim_{x \to 1^+} (-2x - 3) = -5. \][/tex]
Since [tex]\( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \)[/tex], the limits from both sides do not equal [tex]\( f(1) \)[/tex]. Therefore, the function is not continuous at [tex]\( x = 1\)[/tex].
### Graph of f(x)
To complete the graphical inspection:
- Plot [tex]\( 1 - x^2 \)[/tex] for [tex]\( x < 1 \)[/tex].
- Plot [tex]\( -2x - 3 \)[/tex] for [tex]\( x \geq 1 \)[/tex].
You will notice a discontinuity at [tex]\( x = 1 \)[/tex]. This is evident since there is a jump from [tex]\( 0 \)[/tex] (approaching from the left) to [tex]\( -5 \)[/tex] (evaluating at [tex]\( x = 1) \)[/tex].
In conclusion, the function [tex]\( f(x) \)[/tex] defined as:
[tex]\[ f(x) = \begin{cases} 1 - x^2 & \text{if } x < 1 \\ -2x - 3 & \text{if } x \geq 1 \end{cases} \][/tex]
is not continuous at [tex]\( x = 1 \)[/tex] because the left-hand limit and the right-hand limit at [tex]\( x = 1 \)[/tex] are not equal.
The function is given as follows:
[tex]\[ f(x) = \begin{cases} 1 - x^2 & \text{if } x < 1 \\ -2x - 3 & \text{if } x \geq 1 \end{cases} \][/tex]
### Step 1: Graph the function
To graph the function, we need to do the following:
1. Graph the piece [tex]\( 1 - x^2 \)[/tex] for [tex]\( x < 1 \)[/tex]:
This is a downward-opening parabola. When [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex], and it opens downward with its vertex at [tex]\( (0,1) \)[/tex].
2. Graph the piece [tex]\( -2x - 3 \)[/tex] for [tex]\( x \geq 1 \)[/tex]:
This is a linear function with a slope of -2 and a y-intercept of -3. For [tex]\( x = 1 \)[/tex], [tex]\( y = -2(1) - 3 = -5 \)[/tex].
### Step 2: Determine boundaries and transitions
To better visualize the function's continuity at [tex]\( x = 1 \)[/tex], let's calculate a few critical points:
For [tex]\( x < 1 \)[/tex]:
[tex]\[ \begin{aligned} f(0) &= 1 - 0^2 = 1 \\ f(0.5) &= 1 - (0.5)^2 = 1 - 0.25 = 0.75 \\ f(0.9) &= 1 - (0.9)^2 = 1 - 0.81 = 0.19 \end{aligned} \][/tex]
For continuity at [tex]\( x = 1 \)[/tex]:
[tex]\[ \begin{aligned} \lim_{x \to 1^-} f(x) &= f(1^-)= 1 - 1^2 = 0 \\ f(1) &= -2(1) - 3 = -5 \\ \lim_{x \to 1^+} f(x) = f(1^+) &= -2(1 + \epsilon) - 3= -2 - 3 = -5, \end{aligned} \][/tex]
where+ [tex]\(\epsilon\)[/tex] is a small number approaching 0.
### Step 3: Summary and Conclusion
So, let's evaluate:
- From the left-hand side as [tex]\( x \)[/tex] approaches 1 ([tex]\( x<1 \)[/tex]), the limit is:
[tex]\[ \lim_{x \to 1^-} (1 - x^2) = 0. \][/tex]
- From the right-hand side ([tex]\( x \ge 1 \)[/tex]), the limit is:
[tex]\[ \lim_{x \to 1^+} (-2x - 3) = -5. \][/tex]
Since [tex]\( \lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x) \)[/tex], the limits from both sides do not equal [tex]\( f(1) \)[/tex]. Therefore, the function is not continuous at [tex]\( x = 1\)[/tex].
### Graph of f(x)
To complete the graphical inspection:
- Plot [tex]\( 1 - x^2 \)[/tex] for [tex]\( x < 1 \)[/tex].
- Plot [tex]\( -2x - 3 \)[/tex] for [tex]\( x \geq 1 \)[/tex].
You will notice a discontinuity at [tex]\( x = 1 \)[/tex]. This is evident since there is a jump from [tex]\( 0 \)[/tex] (approaching from the left) to [tex]\( -5 \)[/tex] (evaluating at [tex]\( x = 1) \)[/tex].
In conclusion, the function [tex]\( f(x) \)[/tex] defined as:
[tex]\[ f(x) = \begin{cases} 1 - x^2 & \text{if } x < 1 \\ -2x - 3 & \text{if } x \geq 1 \end{cases} \][/tex]
is not continuous at [tex]\( x = 1 \)[/tex] because the left-hand limit and the right-hand limit at [tex]\( x = 1 \)[/tex] are not equal.
