UW MADISON – MATH DEPARTMENT
MATH 211 – SPRING 2017
WORKSHEET 5 - FEB 1, 2017
NAME:
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Review:
• Power Functions: f (x) = x, x 2 , x 3 , x 4 ,
p
x,
p
3
x, x 2/3 , x −1 ;
• Exponential functions f (x) = a x
• Exponential beats polynomial;
x
• Laws of exponents: a x · a y = a x+ y , aa y = a x− y , (a x ) y = a x y , (ab) x = a x b x ;
• Logarithm function f (x) = ln x
• Property of logarithm: ln b = c ⇔ e c = b, ln(e x ) = x, eln x = x;
• Laws of logarithms: ln(x y) = ln x + ln y, ln( xy ) = ln x − ln y, ln(x r ) = r ln x.
• Graphing of exponential and logarithm functions
– Notice the “special point" and asymptote;
– Determine the base is 0 < a < 1 or a > 1;
1. Let’s check the symmetric property (odd or even) of f (x) = x 3 ,
tively.
p
x,
p
3
x, x 2/3 , x −1 respec-
Proof.
• f (x) = x 3 , f (−x) = (−x)3 = −x 3 = − f (x), Odd.
• Neither because domain is x > 0.
p
p
p
3
• f (x) = 3 x, f (−x) = −x = − 3 x = − f (x), Odd.
p
Note. Note that x 1/3 and 3 x are different functions (different domains, the former
is only defined for x ≥ 0 but the latter one is defined for all x ∈ R)
p
p
3
• If we understand x 2/3 as x 2 or ( 3 x)2 then it is even as f (−x) = f (x). But in this
case they are different
p
2
3
x 3 6= x 2
thus the former is only defined for x ≥ 0, so neither odd or even.
• f (x) = x −1 , f (−x) = (−x)−1 = −x −1 , Odd.
2. Sketch the graph of the following functions and describe their relation with y = e x :
(a) y = e−x
(b) y = −e x
(c) y = −e−x
(d) y = e x−1
(e) y = e x + 1
Proof. The graphs are as following
(a)
(b)
(a)
(b)
(e)
3. Sketch the graph of the following functions and describe their relation with y = ln x:
(a) y = ln(−x)
(c) y = − ln(−x)
(b) y = − ln x;
(d) y = ln(x − 1);
(e) y = 1 + ln x.
Proof. The graphs are as following
(a)
(b)
(a)
(b)
(e)
4. Simplify the following expressions:
(a) (3x y 4 )2
Proof. (3x y 4 )2 = 9x 2 y 4 .
(b) e2 ln 3
(Result: 9)
Proof. e2 ln 3 = (eln 3 )2 = 32 = 9.
(c) log4 16
(Result: 2)
Proof. log4 16 = log4 (42 ) = 2.
(d) ln e
p
2
(Result:
Proof. ln e
p
2
=
p
2)
p
p
2 ln e = 2
(e) ln x + a ln y − n ln z
(Result: ln(
x ya
z n ))
Proof. ln x + a ln y − n ln z = ln x + ln y a − ln(z n ) = ln(x y a ) − ln(z n ) = ln(
(e)
log3 9
log2 8
Proof.
x ya
zn )
(Result: 23 )
log3 9
log2 8
=
log3 (32 )
log2 (23 )
(f) ln[(x − 4)(2x + 5)]2
=
2
3
(Result: 2[ln(x − 4) + ln(2x + 5)])
Proof. ln[(x − 4)(2x + 5)]2 = 2[ln(x − 4)(2x + 5)] = 2[ln(x − 4) + ln(2x + 5)].