a. y = ∣ x ∣ y = \sqrt {|x|} y = ∣ x ∣
Solution:
square root expression must be non-negative, so ∣ x ∣ ≥ 0 |x| \ge 0 ∣ x ∣ ≥ 0 - performed for any x ∈ ( − ∞ ; + ∞ ) x \in \left( { - \infty ; + \infty } \right) x ∈ ( − ∞ ; + ∞ ) . Then D ( y ) : x ∈ ( − ∞ ; + ∞ ) D(y):\,x \in \left( { - \infty ; + \infty } \right) D ( y ) : x ∈ ( − ∞ ; + ∞ ) .
Plot the function:
Function is on to if ∀ y ∈ Y ∃ x ∈ X : f ( x ) = y \forall y \in Y\exists x \in X:\,f(x) = y ∀ y ∈ Y ∃ x ∈ X : f ( x ) = y , so, we have onto function.
b. y = 1 − 2 x − x 2 y = 1 - 2x - {x^2} y = 1 − 2 x − x 2
Solution:
There is no restriction for the variable x, so D ( y ) : x ∈ ( − ∞ ; + ∞ ) D(y):\,x \in \left( { - \infty ; + \infty } \right) D ( y ) : x ∈ ( − ∞ ; + ∞ ) .
The function graph is a parabola. Let's find the coordinates of the vertex:
x 0 = − b 2 a = − − 2 − 2 = − 1 ⇒ y 0 = 1 + 2 − 1 = 2 {x_0} = - \frac{b}{{2a}} = - \frac{{ - 2}}{{ - 2}} = - 1 \Rightarrow {y_0} = 1 + 2 - 1 = 2 x 0 = − 2 a b = − − 2 − 2 = − 1 ⇒ y 0 = 1 + 2 − 1 = 2
Find the zeros of the function:
x 2 + 2 x − 1 = 0 {x^2} + 2x - 1 = 0 x 2 + 2 x − 1 = 0
D = 4 + 4 = 8 D = 4 + 4 = 8 D = 4 + 4 = 8
x 1 = − 2 − 8 2 = − 1 − 2 , x 2 = − 1 + 2 {x_1} = \frac{{ - 2 - \sqrt 8 }}{2} = - 1 - \sqrt 2 ,\,\,{x_2} = - 1 + \sqrt 2 x 1 = 2 − 2 − 8 = − 1 − 2 , x 2 = − 1 + 2
We have the graph:
similarly, we have onto function.
Comments