Question #177491

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

one-one or onto.

a. √|𝑥|

b. 1−2𝑥−𝑥2


1
Expert's answer
2021-04-15T07:40:05-0400

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

Solution:

square root expression must be non-negative, so x0|x| \ge 0 - performed for any x(;+)x \in \left( { - \infty ; + \infty } \right). Then D(y):x(;+)D(y):\,x \in \left( { - \infty ; + \infty } \right) .

Plot the function:



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

b. y=12xx2y = 1 - 2x - {x^2}

Solution:

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

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

x0=b2a=22=1y0=1+21=2{x_0} = - \frac{b}{{2a}} = - \frac{{ - 2}}{{ - 2}} = - 1 \Rightarrow {y_0} = 1 + 2 - 1 = 2

Find the zeros of the function:

x2+2x1=0{x^2} + 2x - 1 = 0

D=4+4=8D = 4 + 4 = 8

x1=282=12,x2=1+2{x_1} = \frac{{ - 2 - \sqrt 8 }}{2} = - 1 - \sqrt 2 ,\,\,{x_2} = - 1 + \sqrt 2

We have the graph:




similarly, we have onto function.


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