Respuesta :
Answer:
the function is continuous from the left at x=1 and continuous from the right at x=0
Step-by-step explanation:
a function is continuous from the right , when
when x→a⁺ lim f(x)=f(a)
and from the left when
when x→a⁻ lim f(x)=f(a)
then since the functions presented are continuous , we have to look for discontinuities only when the functions change
for x=0
when x→0⁺ lim f(x)=lim e^x = e^0 = 1
when x→0⁻ lim f(x)=lim (x+4) = (0+4) = 4
then since f(0) = e^0=1 , the function is continuous from the right at x=0
for x=1
when x→1⁺ lim f(x)=lim (8-x) = (8-0) = 8
when x→1⁻ lim f(x)=lim e^x = e^1 = e
then since f(1) = e^1=e , the function is continuous from the left at x=1
Using continuity concepts, it is found that the function is left-continuous at x = 1.
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A function f(x) is said to be continuous at x = a if:
[tex]\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)[/tex]
- If only [tex]\lim_{x \rightarrow a^{-}} f(x) = f(a)[/tex], the function is left-continuous.
- If only [tex]\lim_{x \rightarrow a^{+}} f(x) = f(a)[/tex], the function is right-continuous.
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The piece-wise definition of the function [tex]f(x)[/tex] is:
[tex]x + 4, x < 0[/tex]
[tex]x, 0 \leq x \leq 1[/tex]
[tex]8 - x, x > 1[/tex]
We have to check the continuity at the points in which the definitions change, that is, x = 0 and x = 1.
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At x = 0:
- The definition at 0 is [tex]f(0) = 0[/tex]
- Approaching x = 0 from the left, we have values less than 0, thus:
[tex]\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0} x + 4 = 0 + 4 = 0[/tex]
- Approaching x = 0 from the right, we have values greater than 0, thus:
[tex]\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0} x = 0[/tex]
Since the limits are equal, and also equal to the definition at the point, the function is continuous at x = 0.
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At x = 1:
- The definition at 1 is [tex]f(1) = 1[/tex]
- Approaching x = 1 from the left, we have values less than 1, thus:
[tex]\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1} x = 1[/tex]
- Approaching x = 1 from the right, we have values greater than 1, thus:
[tex]\lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1} 8 - x = 8 - 1 = 7[/tex]
To the right, the limit is different, thus, the function is only left continuous at x = 1.
A similar problem is given at https://brainly.com/question/21447009
