Answer to Question #177491 in Calculus for Talha

Question

Q4. Find the domain and graph the functions. Also tell whether they are

one-one or onto.

a. √|𝑥|

b. 1−2𝑥−𝑥2

Answer 1

a. $y = \sqrt {|x|}$

Solution:

square root expression must be non-negative, so $|x| \ge 0$ - performed for any $x \in \left( { - \infty ; + \infty } \right)$. Then $D(y):\,x \in \left( { - \infty ; + \infty } \right)$ .

Plot the function:

Function is on to if $\forall y \in Y\exists x \in X:\,f(x) = y$ , so, we have onto function.

b. $y = 1 - 2x - {x^2}$

Solution:

There is no restriction for the variable x, so $D(y):\,x \in \left( { - \infty ; + \infty } \right)$ .

The function graph is a parabola. Let's find the coordinates of the vertex:

${x_0} = - \frac{b}{{2a}} = - \frac{{ - 2}}{{ - 2}} = - 1 \Rightarrow {y_0} = 1 + 2 - 1 = 2$

Find the zeros of the function:

${x^2} + 2x - 1 = 0$

$D = 4 + 4 = 8$

${x_1} = \frac{{ - 2 - \sqrt 8 }}{2} = - 1 - \sqrt 2 ,\,\,{x_2} = - 1 + \sqrt 2$

We have the graph:

similarly, we have onto function.