Evaluate The Function K ( X ) = X + 1 K(x)=\sqrt{x+1} K ( X ) = X + 1 ​ For The Given Values Of X X X . Write Exact Answers In Simplified Form. If The Function Is Undefined At The Given Value Of X X X , Use Undefined.(a) K ( 0 K(0 K ( 0 ](b) K ( − 1 K(-1 K ( − 1 ](c)

Evaluating the Function $k(x)=\sqrt{x+1}$ for Given Values of $x$

In this article, we will evaluate the function $k(x)=\sqrt{x+1}$ for the given values of $x$. We will simplify the expressions and provide exact answers. If the function is undefined at the given value of $x$, we will use "Undefined" to indicate this.

Evaluating $k(0)$

To evaluate $k(0)$, we substitute $x=0$ into the function $k(x)=\sqrt{x+1}$.

$k(0) = \sqrt{0+1}$ $k(0) = \sqrt{1}$ $k(0) = 1$

Therefore, $k(0) = 1$.

Evaluating $k(-1)$

To evaluate $k(-1)$, we substitute $x=-1$ into the function $k(x)=\sqrt{x+1}$.

$k(-1) = \sqrt{-1+1}$ $k(-1) = \sqrt{0}$ $k(-1) = 0$

Therefore, $k(-1) = 0$.

Evaluating $k(x)$ for $x<-1$

To evaluate $k(x)$ for $x<-1$, we need to consider the domain of the function $k(x)=\sqrt{x+1}$. The expression inside the square root must be non-negative, so we have $x+1 \geq 0$. Solving for $x$, we get $x \geq -1$. Therefore, the function $k(x)$ is undefined for $x<-1$.

In this article, we evaluated the function $k(x)=\sqrt{x+1}$ for the given values of $x$. We simplified the expressions and provided exact answers. If the function is undefined at the given value of $x$, we used "Undefined" to indicate this. We found that $k(0) = 1$, $k(-1) = 0$, and $k(x)$ is undefined for $x<-1$.

The function $k(x)=\sqrt{x+1}$ is a square root function, which means that the expression inside the square root must be non-negative. This implies that the domain of the function is $x \geq -1$. For values of $x$ less than $-1$, the function is undefined.

To analyze the function $k(x)=\sqrt{x+1}$, we can use the concept of limits. We can evaluate the limit of the function as $x$ approaches $-1$ from the left and the right.

$\lim_{x \to -1^-} k(x) = \lim_{x \to -1^-} \sqrt{x+1} = \lim_{x \to -1^-} \sqrt{(-1)+1} = \lim_{x \to -1^-} \sqrt{0} = 0$

$\lim_{x \to -1^+} k(x) = \lim_{x \to -1^+} \sqrt{x+1} = \lim_{x \to -1^+} \sqrt{(-1)+1} = \lim_{x \to -1^+} \sqrt{0} = 0$

Since the limits from the left and the right are equal, we can conclude that the function $k(x)=\sqrt{x+1}$ is continuous at $x=-1$.

To analyze the function $k(x)=\sqrt{x+1}$ graphically, we can plot the graph of the function. The graph of the function is a square root curve that opens upwards. The graph passes through the point $(0,1)$ and has a vertical asymptote at $x=-1$.

In this article, we evaluated the function $k(x)=\sqrt{x+1}$ for the given values of $x$. We simplified the expressions and provided exact answers. If the function is undefined at the given value of $x$, we used "Undefined" to indicate this. We found that $k(0) = 1$, $k(-1) = 0$, and $k(x)$ is undefined for $x<-1$. We also analyzed the function mathematically and graphically.
Evaluating the Function $k(x)=\sqrt{x+1}$ for Given Values of $x$: Q&A

In our previous article, we evaluated the function $k(x)=\sqrt{x+1}$ for the given values of $x$. We simplified the expressions and provided exact answers. If the function is undefined at the given value of $x$, we used "Undefined" to indicate this. In this article, we will answer some frequently asked questions about the function $k(x)=\sqrt{x+1}$.

Q: What is the domain of the function $k(x)=\sqrt{x+1}$?

A: The domain of the function $k(x)=\sqrt{x+1}$ is $x \geq -1$. This is because the expression inside the square root must be non-negative, so we have $x+1 \geq 0$. Solving for $x$, we get $x \geq -1$.

Q: Is the function $k(x)=\sqrt{x+1}$ continuous at $x=-1$?

A: Yes, the function $k(x)=\sqrt{x+1}$ is continuous at $x=-1$. We can evaluate the limit of the function as $x$ approaches $-1$ from the left and the right.

$\lim_{x \to -1^-} k(x) = \lim_{x \to -1^-} \sqrt{x+1} = \lim_{x \to -1^-} \sqrt{(-1)+1} = \lim_{x \to -1^-} \sqrt{0} = 0$

$\lim_{x \to -1^+} k(x) = \lim_{x \to -1^+} \sqrt{x+1} = \lim_{x \to -1^+} \sqrt{(-1)+1} = \lim_{x \to -1^+} \sqrt{0} = 0$

Since the limits from the left and the right are equal, we can conclude that the function $k(x)=\sqrt{x+1}$ is continuous at $x=-1$.

Q: What is the value of $k(0)$?

A: The value of $k(0)$ is $1$. We can substitute $x=0$ into the function $k(x)=\sqrt{x+1}$ to get $k(0) = \sqrt{0+1} = \sqrt{1} = 1$.

Q: What is the value of $k(-1)$?

A: The value of $k(-1)$ is $0$. We can substitute $x=-1$ into the function $k(x)=\sqrt{x+1}$ to get $k(-1) = \sqrt{-1+1} = \sqrt{0} = 0$.

Q: Is the function $k(x)=\sqrt{x+1}$ defined for $x<-1$?

A: No, the function $k(x)=\sqrt{x+1}$ is not defined for $x<-1$. This is because the expression inside the square root must be non-negative, so we have $x+1 \geq 0$. Solving for $x$, we get $x \geq -1$. Therefore, the function $k(x)$ is undefined for $x<-1$.

Q: What is the graph of the function $k(x)=\sqrt{x+1}$?

A: The graph of the function $k(x)=\sqrt{x+1}$ is a square root curve that opens upwards. The graph passes through the point $(0,1)$ and has a vertical asymptote at $x=-1$.

In this article, we answered some frequently asked questions about the function $k(x)=\sqrt{x+1}$. We discussed the domain of the function, its continuity at $x=-1$, and its values at $x=0$ and $x=-1$. We also analyzed the graph of the function. We hope that this article has been helpful in understanding the function $k(x)=\sqrt{x+1}$.