If $f\left(x\right)=\left\{\begin{array}{l}1+xifx<0\\ x^{2}ifx\ge 0\end{array}\right.$

(a) Find the domain of each function

(b) Locate any intercepts

(c) Graph each function

(d) Based on the graph,Find the range

(e) Is f continuous on its domain?

The given function$f\left(x\right)=\left\{\begin{array}{l}1+xifx<0\\ x^{2}ifx\ge 0\end{array}\right.$

The domain of the function $f\left(x\right)$ , is the set of all the possible values of x.

The value of the function $f\left(x\right)$when$x<0$ is given by 1+x .
In the expression $1+x,1$ is added to x. This operation can be performed on any real number.

The value of the function $f\left(x\right)$ when $x\ge 1$ is given by x².
In the expression x² , the value of the variablex is raised to the power of 2 . These operations can be performed on any real number.
So, the domain ofx is the set of all real numbers.
Therefore, the domain of the given function is the set of all the real numbers.

The x-intercepts are those points for which the y -coordinate is zero and the y-intercepts are those points for which the x-coordinate is zero.

Determine the points on the graph for which thex-coordinate is 0.
The value of the function $f\left(x\right)$ or the y-coordinate when $x=0$ is given by x².

$$\begin{gathered} f\left(0\right)=0^2 \\ f\left(0\right)=0 \end{gathered}$$

So,the y-intercept is 0.

Determine the points on the graph for which they -coordinate is 0 .
Consider the function $f\left(x\right)=1+x$ . If the y-coordinate is 0 , then $f\left(x\right)=0$.

$$\begin{gathered} 0=1+x \\ x=-1 \end{gathered}$$

So, thex -intercept is -1 .
Therefore the intercepts are (-1,0) and (0,0)

For plotting the graph of the parabola $y=x^2$, in the part for which $x\ge 0$ , substitute various values for x in the equation to obtain some points on the graph.

Plot the points and draw the line to get the graph of the function.

From the points plotted on the graph we can see that for each numbery , there is at least one number x in the domain.
So, the range off is the set of all the real numbers.

The only point at which the function might have behaved in a manner that it becomes discontinuous is $x=0$. But at this point, the value of the function from the left of 0 and the right of 0 are given as:

$$\begin{gathered} f\left(0^{-}\right)=1+x \\ =1+0 \\ =1 \\ and \\ f\left(0^{+}\right)=x^2 \\ =0^2 \\ =0 \end{gathered}$$

Hence, $f\left(0^{-}\right)\ne f\left(0^{+}\right)$

So at the break point the function is discontinuous.
Hence the function is discontinuous in its domain only at the point$x=0$ .
Also from the graph it can be clearly seen that the function is discontinuous at the point $x=0$.