a. y=∣x∣
Solution:
square root expression must be non-negative, so ∣x∣≥0 - performed for any x∈(−∞;+∞). Then D(y):x∈(−∞;+∞) .
Plot the function:
Function is on to if ∀y∈Y∃x∈X:f(x)=y , so, we have onto function.
b. y=1−2x−x2
Solution:
There is no restriction for the variable x, so D(y):x∈(−∞;+∞) .
The function graph is a parabola. Let's find the coordinates of the vertex:
x0=−2ab=−−2−2=−1⇒y0=1+2−1=2
Find the zeros of the function:
x2+2x−1=0
D=4+4=8
x1=2−2−8=−1−2,x2=−1+2
We have the graph:
similarly, we have onto function.
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