# 1 In Class Questions MATH 151-Fall 02 October 24

advertisement ```1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
October 24
1. What is the definition of one to one, 1-1?
f (a) 6= f (b), whenever a 6= b
or
If a 6= b, then f (a) 6= f (b)
or
If f (a) = f (b), then a = b
or
a = b whenever f (a) = f (b)
2. Is y = (x − 1)8 1-1? If not, could it be made to be 1-1?
The function is not 1-1 since it has a degree that is even. If we restrict the domain to x ≥ 1
then the function will be 1-1.
3. Show that y =
3x + 7
is one to one and then find the inverse.
x+5
If f (a) = f (b) show a = b
3a + 7
3b + 7
=
a+5
b+5
(3a + 7)(b + 5) = (3b + 7)(a + 5)
3ab + 15a + 7b + 35 = 3ab + 15b + 7a + 35
8a = 8b
a=b
Hence the function is 1-1.
To find the inverse:
3y + 7
x=
y+5
x(y + 5) = 3y + 7
xy + 5x = 3y + 7
xy − 3y = 7 − 5x
y(x − 3) = 7 − 5x
y=
7 − 5x
x−3
4. Derive the formula for taking the derivative of an inverse function at x = a.
If g(x) and f (x) are inverses of each other then f (g(x)) = x.
now take a derivative and solve for g0 (x) and plug in x = a
f 0 (g(x)) ∗ g0 (x) = 1
g0 (a) =
1
f 0 (g(a))
as long as f 0 (g(a)) 6= 0
5. If g(x) is the inverse of f (x), find the indicated derivative.
(a) g0 (3) if f (x) = 5 + 2x + e2x+3
Find 2 such that
g(3) = 2
f (2) = 3
If we plug in a −1.5 into f (x) we get an answer of 3.
now f 0 (x) = 2 + 2e2x+3
1
1
1
1
g0 (3) = 0
=
= 0
=
2∗−1.5+3
f (g(3))
f (−1.5)
2 + 2e
4
x
−
2
(b) g0 (−3) if f (x) =
x+2
Find 2 such that
g(−3) = 2
151 WebCalc Fall 2002-copyright Joe Kahlig
f (2) = −3
If we plug in a −1 into f (x) we get an answer of −3.
x + 2 − (x − 2)
now f 0 (x) =
(x + 2)2
1
1
1
g0 (−3) = 0
= 0
=
f (g(−3))
f (−1)
4
0
0
(c) g (9) if f (10) = 9, f (9) = 3, f (9) = 7, and f 0 (10) = 4
1
1
1
g0 (9) = 0
= 0
=
f (g(9))
f (10)
4
2
```