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To analyze the function [tex]\( f(x) \)[/tex] and determine its continuity, follow these steps:
1. Define the pieces of the function:
- For [tex]\(-4 \leq x < 4\)[/tex], [tex]\( f(x) = -10 + x^2 \)[/tex].
- For [tex]\( x \geq 4\)[/tex], [tex]\( f(x) = 12 - x \)[/tex].
2. Analyze and graph each piece:
- Piece 1: [tex]\(-10 + x^2\)[/tex] on [tex]\([-4, 4)\)[/tex]
- This is a parabola that opens upwards, shifted down by 10 units.
- At [tex]\( x = -4 \)[/tex], [tex]\( f(x) = -10 + (-4)^2 = 6 \)[/tex].
- As [tex]\( x \)[/tex] approaches 4 from the left, [tex]\( f(x) \)[/tex] gets closer to [tex]\( 6 \)[/tex] from above.
- Piece 2: [tex]\( 12 - x \)[/tex] for [tex]\( x \geq 4 \)[/tex]
- This is a linear function with a negative slope.
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 12 - 4 = 8 \)[/tex].
- As [tex]\( x \)[/tex] increases, the value of [tex]\( f(x) \)[/tex] decreases.
3. Determine the values at the endpoints and any possible discontinuity:
- At [tex]\( x = 4 \)[/tex]:
- From the left ([tex]\( x \to 4^- \)[/tex]), [tex]\( f(x) = -10 + 4^2 = 6 \)[/tex].
- From the right ([tex]\( x \to 4^+ \)[/tex]), [tex]\( f(x) = 12 - 4 = 8 \)[/tex].
- The value of [tex]\( f(4) \)[/tex] is 8 since [tex]\( 12 - x \)[/tex] is defined for [tex]\( x \geq 4 \)[/tex].
4. Plotting the function:
- Draw the parabola [tex]\(-10 + x^2\)[/tex] for [tex]\(-4 \leq x < 4\)[/tex]. The curve starts at [tex]\( ( -4, 6 ) \)[/tex] and gets close to but not including [tex]\( (4, 6) \)[/tex].
- Draw the line [tex]\( 12 - x \)[/tex] starting at [tex]\( (4, 8) \)[/tex] and extending for [tex]\( x \geq 4 \)[/tex].
The graph of the function should look something like this:
[tex]\[ \begin{array}{ccc} \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = $x$, ylabel = $f(x)$, ymin=-10, ymax=15, xmin=-5, xmax=10, xtick={-4,-2,0,2,4,6,8}, ytick={-10,-5,0,5,10,12,14}, extra x ticks={4}, extra x tick labels={$4$}, extra y ticks={8}, extra y tick labels={$8$}, ] \addplot[ domain=-4:4, samples=100, color=blue, ] {-10 + x^2}; \addplot[ domain=4:10, samples=100, color=red, ] {12 - x}; \node at (axis cs: -4,6) [circle,fill,inner sep=1.5pt, label=left:{$( -4, 6 )$}] {}; \node at (axis cs: 4,8) [circle,fill,inner sep=1.5pt, label=above right:{$( 4, 8 )$}] {}; \node at (axis cs: 4,6) [circle,draw,fill=white,inner sep=1.5pt, label=below left:{$( 4, 6 )$}] {}; \end{axis} \end{tikzpicture} \end{array} \][/tex]
5. Check continuity:
- Continuity at [tex]\( x = 4 \)[/tex]:
- The left-hand limit as [tex]\( x \)[/tex] approaches 4 is [tex]\( \lim_{{x \to 4^-}} f(x) = 6 \)[/tex].
- The right-hand limit as [tex]\( x \)[/tex] approaches 4 is [tex]\( \lim_{{x \to 4^+}} f(x) = 8 \)[/tex].
- Since the limits from the left and right do not match ([tex]\( 6 \neq 8 \)[/tex]), the function [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex].
In conclusion, [tex]\( f(x) \)[/tex] is not continuous because the function has a jump discontinuity at [tex]\( x = 4 \)[/tex]. The graph visually illustrates this discontinuity with a filled circle at [tex]\( (4, 8) \)[/tex] and an open circle at [tex]\( (4, 6) \)[/tex].
1. Define the pieces of the function:
- For [tex]\(-4 \leq x < 4\)[/tex], [tex]\( f(x) = -10 + x^2 \)[/tex].
- For [tex]\( x \geq 4\)[/tex], [tex]\( f(x) = 12 - x \)[/tex].
2. Analyze and graph each piece:
- Piece 1: [tex]\(-10 + x^2\)[/tex] on [tex]\([-4, 4)\)[/tex]
- This is a parabola that opens upwards, shifted down by 10 units.
- At [tex]\( x = -4 \)[/tex], [tex]\( f(x) = -10 + (-4)^2 = 6 \)[/tex].
- As [tex]\( x \)[/tex] approaches 4 from the left, [tex]\( f(x) \)[/tex] gets closer to [tex]\( 6 \)[/tex] from above.
- Piece 2: [tex]\( 12 - x \)[/tex] for [tex]\( x \geq 4 \)[/tex]
- This is a linear function with a negative slope.
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 12 - 4 = 8 \)[/tex].
- As [tex]\( x \)[/tex] increases, the value of [tex]\( f(x) \)[/tex] decreases.
3. Determine the values at the endpoints and any possible discontinuity:
- At [tex]\( x = 4 \)[/tex]:
- From the left ([tex]\( x \to 4^- \)[/tex]), [tex]\( f(x) = -10 + 4^2 = 6 \)[/tex].
- From the right ([tex]\( x \to 4^+ \)[/tex]), [tex]\( f(x) = 12 - 4 = 8 \)[/tex].
- The value of [tex]\( f(4) \)[/tex] is 8 since [tex]\( 12 - x \)[/tex] is defined for [tex]\( x \geq 4 \)[/tex].
4. Plotting the function:
- Draw the parabola [tex]\(-10 + x^2\)[/tex] for [tex]\(-4 \leq x < 4\)[/tex]. The curve starts at [tex]\( ( -4, 6 ) \)[/tex] and gets close to but not including [tex]\( (4, 6) \)[/tex].
- Draw the line [tex]\( 12 - x \)[/tex] starting at [tex]\( (4, 8) \)[/tex] and extending for [tex]\( x \geq 4 \)[/tex].
The graph of the function should look something like this:
[tex]\[ \begin{array}{ccc} \begin{tikzpicture} \begin{axis}[ axis lines = middle, xlabel = $x$, ylabel = $f(x)$, ymin=-10, ymax=15, xmin=-5, xmax=10, xtick={-4,-2,0,2,4,6,8}, ytick={-10,-5,0,5,10,12,14}, extra x ticks={4}, extra x tick labels={$4$}, extra y ticks={8}, extra y tick labels={$8$}, ] \addplot[ domain=-4:4, samples=100, color=blue, ] {-10 + x^2}; \addplot[ domain=4:10, samples=100, color=red, ] {12 - x}; \node at (axis cs: -4,6) [circle,fill,inner sep=1.5pt, label=left:{$( -4, 6 )$}] {}; \node at (axis cs: 4,8) [circle,fill,inner sep=1.5pt, label=above right:{$( 4, 8 )$}] {}; \node at (axis cs: 4,6) [circle,draw,fill=white,inner sep=1.5pt, label=below left:{$( 4, 6 )$}] {}; \end{axis} \end{tikzpicture} \end{array} \][/tex]
5. Check continuity:
- Continuity at [tex]\( x = 4 \)[/tex]:
- The left-hand limit as [tex]\( x \)[/tex] approaches 4 is [tex]\( \lim_{{x \to 4^-}} f(x) = 6 \)[/tex].
- The right-hand limit as [tex]\( x \)[/tex] approaches 4 is [tex]\( \lim_{{x \to 4^+}} f(x) = 8 \)[/tex].
- Since the limits from the left and right do not match ([tex]\( 6 \neq 8 \)[/tex]), the function [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 4 \)[/tex].
In conclusion, [tex]\( f(x) \)[/tex] is not continuous because the function has a jump discontinuity at [tex]\( x = 4 \)[/tex]. The graph visually illustrates this discontinuity with a filled circle at [tex]\( (4, 8) \)[/tex] and an open circle at [tex]\( (4, 6) \)[/tex].
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