Question

Find each x-value at which f is discontinuous and for each x-value, determine whether f is continuous from the right, or from the left, or neither.
f(x) = x + 9 if x < 0
e^x if 0 ≤ x ≤ 1
4 − x if x > 1
x = (smaller value)
continuous from the right, or continuous from the left, or neither
x = (larger value)
continuous from the right, or continuous from the left, or neither

Answer 1

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:

$\lim_{x \rightarrow a^{-}} f(x) = \lim_{x \rightarrow a^{+}} f(x) = f(a)$

• If $\lim_{x \rightarrow a^{-}} f(x) = f(a)$, the function is left-continuous.
• If $\lim_{x \rightarrow a^{+}} f(x) = f(a)$, 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:

$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} x + 9 = 0 + 9 = 9$

• Approaching from the right, it is more than 0, thus:

$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} e^x = e^0 = 1$

• The numeric value is:

$f(x) = e^0 = 1$

• Since $[tex]\lim_{x \rightarrow 0^+} f(x) = f(0)$, the function is right-continuous at x = 0.

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At x = 1.

$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1} e^x = e^1 = e$

$\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1} 4 - x = 4 - x = 3$

$f(1) = e^1 = e$

• Since $[tex]\lim_{x \rightarrow 1^-} f(x) = f(1)$, the function is left-continuous at x = 0.

A similar problem is given at https://brainly.com/question/21447009