QuestionFind the values of $x \in \mathbb{R}$ for which $f(x)=\frac{x^{2}+1}{x-1}$ is defined and continuous (in interval notation).

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\frac{x^{}+1}{x-1}$ . We are looking for values of $x$ in the real numbers ( $\mathbb{R}$ ) for which the function is both defined and continuous.

STEP 2

A function is defined for all real numbers except where the denominator is zero. So, we first need to find the values of $x$ for which the denominator of the function is zero.
$x-1=0$

STEP 3

olve the equation $x-1=0$ for $x$ .
$x=1$

STEP 4

So, the function is undefined at $x=1$ .

STEP 5

A function is continuous everywhere it is defined, except possibly at the points where it is undefined. So, the function is continuous for all real numbers except possibly at $x=1$ .

STEP 6

To check if the function is continuous at $x=1$ , we need to check the limit of the function as $x$ approaches1 from both the left and the right. If these two limits are equal and finite, then the function is continuous at $x=1$ .

STEP 7

First, check the limit as $x$ approaches1 from the left.
$\lim{{x \to1^-}} \frac{x^{2}+1}{x-1}$

STEP 8

As $x$ approaches1 from the left, the denominator approaches0 from the negative side, causing the fraction to approach negative infinity.

STEP 9

Next, check the limit as $x$ approaches from the right.
$\lim{{x \to^+}} \frac{x^{2}+}{x-}$

STEP 10

As $x$ approaches from the right, the denominator approaches0 from the positive side, causing the fraction to approach positive infinity.

STEP 11

Since the two limits as $x$ approaches from the left and the right are not equal, the function is not continuous at $x=$ .

STEP 12

Therefore, the function is both defined and continuous for all real numbers except $x=$ .
The solution is $x \in \mathbb{R} \setminus \{\}$ , or in interval notation, $(-\infty,) \cup (, \infty)$ .

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QuestionFind the values of $x \in \mathbb{R}$ for which $f(x)=\frac{x^{2}+1}{x-1}$ is defined and continuous (in interval notation).

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=\frac{x^{}+1}{x-1}$ . We are looking for values of $x$ in the real numbers ( $\mathbb{R}$ ) for which the function is both defined and continuous.

STEP 2

A function is defined for all real numbers except where the denominator is zero. So, we first need to find the values of $x$ for which the denominator of the function is zero.
$x-1=0$

STEP 3

olve the equation $x-1=0$ for $x$ .
$x=1$

STEP 4

So, the function is undefined at $x=1$ .

STEP 5

A function is continuous everywhere it is defined, except possibly at the points where it is undefined. So, the function is continuous for all real numbers except possibly at $x=1$ .

STEP 6

To check if the function is continuous at $x=1$ , we need to check the limit of the function as $x$ approaches1 from both the left and the right. If these two limits are equal and finite, then the function is continuous at $x=1$ .

STEP 7

First, check the limit as $x$ approaches1 from the left.
$\lim{{x \to1^-}} \frac{x^{2}+1}{x-1}$

STEP 8

As $x$ approaches1 from the left, the denominator approaches0 from the negative side, causing the fraction to approach negative infinity.

STEP 9

Next, check the limit as $x$ approaches from the right.
$\lim{{x \to^+}} \frac{x^{2}+}{x-}$

STEP 10

As $x$ approaches from the right, the denominator approaches0 from the positive side, causing the fraction to approach positive infinity.

STEP 11

Since the two limits as $x$ approaches from the left and the right are not equal, the function is not continuous at $x=$ .

STEP 12

Therefore, the function is both defined and continuous for all real numbers except $x=$ .
The solution is $x \in \mathbb{R} \setminus \{\}$ , or in interval notation, $(-\infty,) \cup (, \infty)$ .

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QuestionFind the values of x∈Rx \in \mathbb{R}x∈R for which f(x)=x2+1x−1f(x)=\frac{x^{2}+1}{x-1}f(x)=x−1x2+1​ is defined and continuous (in interval notation).

Studdy Solution

STEP 1

Assumptions1. The function is f(x)=x+1x−1f(x)=\frac{x^{}+1}{x-1}f(x)=x−1x+1​ . We are looking for values of xxx in the real numbers (R\mathbb{R}R) for which the function is both defined and continuous.

STEP 2

A function is defined for all real numbers except where the denominator is zero. So, we first need to find the values of xxx for which the denominator of the function is zero.x−1=0x-1=0x−1=0

STEP 3

olve the equation x−1=0x-1=0x−1=0 for xxx.x=1x=1x=1

STEP 4

So, the function is undefined at x=1x=1x=1.

STEP 5

A function is continuous everywhere it is defined, except possibly at the points where it is undefined. So, the function is continuous for all real numbers except possibly at x=1x=1x=1.

STEP 6

To check if the function is continuous at x=1x=1x=1, we need to check the limit of the function as xxx approaches1 from both the left and the right. If these two limits are equal and finite, then the function is continuous at x=1x=1x=1.

STEP 7

First, check the limit as xxx approaches1 from the left.lim⁡x→1−x2+1x−1\lim{{x \to1^-}} \frac{x^{2}+1}{x-1}limx→1−x−1x2+1​

STEP 8

As xxx approaches1 from the left, the denominator approaches0 from the negative side, causing the fraction to approach negative infinity.

STEP 9

Next, check the limit as xxx approaches from the right.lim⁡x→+x2+x−\lim{{x \to^+}} \frac{x^{2}+}{x-}limx→+x−x2+​

STEP 10

As xxx approaches from the right, the denominator approaches0 from the positive side, causing the fraction to approach positive infinity.

STEP 11

Since the two limits as xxx approaches from the left and the right are not equal, the function is not continuous at x=x=x=.

STEP 12

Therefore, the function is both defined and continuous for all real numbers except x=x=x=.The solution is x∈R∖{}x \in \mathbb{R} \setminus \{\}x∈R∖{}, or in interval notation, (−∞,)∪(,∞)(-\infty,) \cup (, \infty)(−∞,)∪(,∞).

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QuestionFind the values of $x \in \mathbb{R}$ for which $f(x)=\frac{x^{2}+1}{x-1}$ is defined and continuous (in interval notation).

Assumptions1. The function is $f(x)=\frac{x^{}+1}{x-1}$ . We are looking for values of $x$ in the real numbers ( $\mathbb{R}$ ) for which the function is both defined and continuous.

A function is defined for all real numbers except where the denominator is zero. So, we first need to find the values of $x$ for which the denominator of the function is zero.
$x-1=0$

olve the equation $x-1=0$ for $x$ .
$x=1$

So, the function is undefined at $x=1$ .

A function is continuous everywhere it is defined, except possibly at the points where it is undefined. So, the function is continuous for all real numbers except possibly at $x=1$ .

To check if the function is continuous at $x=1$ , we need to check the limit of the function as $x$ approaches1 from both the left and the right. If these two limits are equal and finite, then the function is continuous at $x=1$ .

First, check the limit as $x$ approaches1 from the left.
$\lim{{x \to1^-}} \frac{x^{2}+1}{x-1}$

As $x$ approaches1 from the left, the denominator approaches0 from the negative side, causing the fraction to approach negative infinity.

Next, check the limit as $x$ approaches from the right.
$\lim{{x \to^+}} \frac{x^{2}+}{x-}$

As $x$ approaches from the right, the denominator approaches0 from the positive side, causing the fraction to approach positive infinity.

Since the two limits as $x$ approaches from the left and the right are not equal, the function is not continuous at $x=$ .

Therefore, the function is both defined and continuous for all real numbers except $x=$ .
The solution is $x \in \mathbb{R} \setminus \{\}$ , or in interval notation, $(-\infty,) \cup (, \infty)$ .