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1. Let

g: \mathbb{R} \rightarrow \mathbb{R}

be the function defined by \[ g(x...

May 27, 2024

Solution by Steps

step 1

To show that

g(x) = x|x|

is continuous at

0

, we need to check the limit of

g(x)

x

approaches

0

from both sides

step 2

For x > 0,

g(x) = x \cdot x = x^2

. For x < 0,

g(x) = x \cdot (-x) = -x^2

step 3

Calculate the limit from the right:

\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} x^2 = 0

step 4

Calculate the limit from the left:

\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} -x^2 = 0

step 5

Since both limits are equal to

0

and

g(0) = 0

g

is continuous at

0

step 6

To check if

g

is continuous everywhere, note that

g(x) = x|x|

is a polynomial function, and polynomial functions are continuous everywhere on

\mathbb{R}

Answer

g

is continuous at

0

and everywhere on

\mathbb{R}

Key Concept

Continuity of a function

Explanation

A function is continuous at a point if the limit from both sides equals the function's value at that point.

Question 2

step 1

To determine the continuity of

g(x)

defined piecewise, we need to check the points where the definition changes:

x = 0

step 2

For

x \leq 0

g(x) = x^2 + 1

. For x > 0,

g(x) = \ln(1 + x)

step 3

Calculate the limit from the left:

\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x^2 + 1) = 1

step 4

Calculate the limit from the right:

\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \ln(1 + x) = \ln(1) = 0

step 5

Since the left and right limits are not equal,

g

is not continuous at

0

step 6

For

x \neq 0

, both

x^2 + 1

and

\ln(1 + x)

are continuous functions

Answer

g

is continuous for

x \neq 0

Key Concept

Piecewise function continuity

Explanation

A piecewise function is continuous at a point where the pieces meet if the limits from both sides are equal and match the function's value at that point.

Question 3

step 1

To determine the continuity of

g(x)

defined piecewise, we need to check the points where the definition changes:

x = 0

and

x = \pi

step 2

For x < 0,

g(x) = e^{-x}

. For

0 \leq x \leq \pi

g(x) = \cos x + 1

. For x > \pi,

g(x) = x - \pi

step 3

Calculate the limit from the left at

x = 0

\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} e^{-x} = 1

step 4

Calculate the limit from the right at

x = 0

\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (\cos x + 1) = 2

step 5

Since the left and right limits at

x = 0

are not equal,

g

is not continuous at

0

step 6

Calculate the limit from the left at

x = \pi

\lim_{x \to \pi^-} g(x) = \lim_{x \to \pi^-} (\cos x + 1) = 0

step 7

Calculate the limit from the right at

x = \pi

\lim_{x \to \pi^+} g(x) = \lim_{x \to \pi^+} (x - \pi) = 0

step 8

Since the left and right limits at

x = \pi

are equal and

g(\pi) = 0

g

is continuous at

\pi

step 9

For

x \neq 0

and

x \neq \pi

, each piece of

g(x)

is continuous

Answer

g

is continuous for

x \neq 0

and

x = \pi

Key Concept

Continuity at piecewise function boundaries

Explanation

A piecewise function is continuous at a boundary point if the limits from both sides are equal and match the function's value at that point.

Question 4

step 1

To determine the continuity of

g(x)

x = 3

, we need to check the limit of

g(x)

x

approaches

3

from both sides

step 2

For

x \neq 3

g(x) = \frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3

step 3

Calculate the limit from both sides:

\lim_{x \to 3} g(x) = \lim_{x \to 3} (x + 3) = 6

step 4

For

g

to be continuous at

x = 3

, we need

g(3) = k = 6

step 5

Therefore,

g

is continuous everywhere if

k = 6

Answer

g

is continuous everywhere if

k = 6

Key Concept

Continuity at a point

Explanation

A function is continuous at a point if the limit from both sides equals the function's value at that point.