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Sagot :
Using continuity concepts, the answers are given by:
- The function is right-continuous at x = 0.
- The function is left-continuous at x = 1.
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A function f(x) is continuous at x = a if:
[tex]\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)[/tex]
- If [tex]\lim_{x \rightarrow a^{-}} f(x) = f(a)[/tex], the function is left-continuous.
- If [tex]\lim_{x \rightarrow a^{+}} f(x) = f(a)[/tex], the function is right-continuous.
- For the given function, the continuity is tested at the points in which the definitions are changed, x = 0 and x = 1.
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At x = 0.
- Approaching from the left, it is less than 0, thus:
[tex]\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} x + 9 = 0 + 9 = 9[/tex]
- Approaching from the right, it is more than 0, thus:
[tex]\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} e^x = e^0 = 1[/tex]
- The numeric value is:
[tex]f(x) = e^0 = 1[/tex]
- Since [tex][tex]\lim_{x \rightarrow 0^+} f(x) = f(0)[/tex], the function is right-continuous at x = 0.
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At x = 1.
[tex]\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} e^x = e^1 = e[/tex]
[tex]\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} 4 - x = 4 - x = 3[/tex]
[tex]f(1) = e^1 = e[/tex]
- Since [tex][tex]\lim_{x \rightarrow 1^-} f(x) = f(1)[/tex], the function is left-continuous at x = 0.
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