Math 211-330/332 Worksheet 2
Jan 25, 2016
Solutions
Review:
• Piecewise functions (different rules in different areas);
• Even function: f(x)=f(-x);
Odd function: f(x)=-f(-x);
• Combination of functions: +, −, ∗ ⇒ intersection of the two domains;
/ ⇒intersection of the two domains and bottom 6= 0;
• Composition of functions : f (g(x)); Notice that f (g(x)) 6= g(f (x));
• Transformation of functions:
y = f (x) + c: shift c units upward;
y = f (x) − c: shift c units downwards;
y = f (x − c): shift c units to the right;
y = f (x + c): shift c units to the left;
y = cf (x), c > 1: stretch vertically by a factor of c;
y = 1c f (x), c > 1: compress vertically by a factor of c;
y = f (cx), c > 1: compress horizontally by a factor of c;
y = f ( 1c x), c > 1: stretch horizontally by a factor of c;
y = −f (x), c > 1: reflect about the x-axis;
y = f (−x), c > 1: reflect about the y-axis.
(Use “WolframAlpha” and play with these transformations.)
1. Let f be the function defined by the requirement that for any x one has
y = f (x) ⇔ y is the largest of all possible solutions of y 2 = 3x2 − 2xy
Find a formula for f . What are the domain and range of f ?
Solution.
y 2 = 3x2 − 2xy ⇒ 3x2 − 2xy − y 2 = 0 ⇒ (3x + y)(x − y) = 0
This give us two possible functions which can satisfy y 2 = 3x2 − 2xy. One is y = x
(from x − y = 0) and the other one is y = −3x (from 3x + y = 0).
From the graph of the two functions, we can see that on (−∞, 0), the red one is larger
than the blue one; on (0, +∞), the blue one is larger than the red one. So we combine
these two into a new piecewise function defined by
(
y=
(red) − 3x x ∈ (−∞, 0)
(blue)x
x ∈ [0, +∞)
We can check that this function satisfy the condition y 2 = 3x2 − 2xy. The domain is
(−∞, +∞) and the range is [0, +∞).
2. Determine whether f is even, odd, or neither.
(a). f (x) =
(b).f (x) =
(c).f (x) =
x
x2 +1
x2
x4 +1
x
x+1
Solution. (a). f (−x) =
(b).f (−x) =
(−x)2
(−x)4 +1
(c). f (−x) =
−x
−x+1
=
−x
(−x)2 +1
x2
x4 +1
= − x2x+1 = −f (x) ⇒ odd function;
= f (x) ⇒ even function;
⇒ neither odd function nor even function.
1
3. If f (x) = x−1
, g(x) =
and the domains.
√
−x2 − x, write a formula for each of the following functions
(a). A(x) = f (x) + g(x);
(b).B(x) = f (x) − g(x);
(c). C(x) = f (x)g(x);
(d). D(x) =
f (x)
g(x) .
Solution. Domain for f (x) is x 6= 1, domain for g(x) is −1 ≤ x ≤ 0.
√
1
(a).A(x) = x−1
+ −x2 − x, domain: −1 ≤ x ≤ 0;
√
1
(b).B(x) = x−1
− −x2 − x, domain: −1 ≤ x ≤ 0;
√
(c).C(x) =
(d).D(x) =
−x2 −x
x−1 , domain: −1 ≤ x ≤ 0;
√1
, domain: −1 < x
(x−1) −x2 −x
< 0.
4. A function f is given that satisfies
f (x + 3) = x2
for all real numbers x.
Compute
(a).f (0)
(b).f (3)
(c).f (π)
(d).f (t)
(e).f (x)
(f).f (f (2))
(g).f (2f (x)).
Solution. (a). Let x + 3 = 0 so that the left hand side becomes f (0). Then we solve
out x = −3, plug in the right hand side. Now we have f (0) = (−3)2 = 9.
It’s OK to do (a) (b) (c) in this way. But we will get trouble for the rest.
So my suggestion to solve this kind of sequence problem: Think of f (g(x)), figure
out what is g(x) and what is f (t).
Step 1: Let t = g(x) = x + 3. So we can get f (t) on the left hand side instead of
f (x + 3);
Step 2: Represent x by t. From Step 1, we have t = x + 3 ⇒ x = t − 3;
Step 3: Plug the representation of x in the right hand side. Then we have
f (t) = f (x + 3) = x2 = (t − 3)2 ⇒ f (t) = (t − 3)2
We know the fact that f (t) = (t − 3)2 is the same thing as f (x) = (x − 3)2 .
Now (a),(b),(c),(d),(e) becomes trivial. (Just plug in.)
(f )
f (2) = (2 − 3)2 = 1
f (f (2)) = f (1) = (1 − 3)2 = 4
(g)
f (x) = (x − 3)2
2f (x) = 2(x − 3)2
f (2f (x)) = f (2(x − 3)2 ) = (2(x − 3)2 − 3)2
Remark: we can also relate this with function transformation.