Page 82 - Calculus Workbook For Dummies

66 Part III: Differentiation

2
In problem 4, you see that the top of this parabola ( y = - x + 5) is at the point (0,5) and that
the derivative is zero there because the parabola is neither going up nor down at its peak. That
explanation was based on common sense. But now, with the result given by the difference quo-
l
tient, namely y = - 2 x, you have a rigorous confirmation of the derivative’s value at (0,5). Just
l
plug 0 in for x in y = - 2 x, and you get y = l 0.
2
j Figure the derivative of g x = 4 x + 5 using the difference quotient. g x =
^ h
l ^ h
4 x + 5
If you got this one, give yourself a pat on the back. It’s a bit tricky.
g x = 4 x + 5
^ h
h +
4 ^ x + h 5 - 4 x + 5
g x = lim
l ^ h
h " 0 h
4 x + 4 h + 5 - 4 x + 5
= lim
h " 0 h
` 4 x + 4 h + 5 - 4 x + 5j ` 4 x + 4 h + 5 + 4 x + 5j
= lim $ _ Conjugate multiplicationi
h " 0 h ` 4 x + 4 h + 5 + 4 x + 5j
4 ^ x + 4 h + 5 - ^h 4 x + 5h
2
= lim ` Because ^ a - h ^ b = a - b 2 , of coursej
b a + h
h " 0
h` 4 x + 4 h + 5 + 4 x + 5j
4 h
= lim
h " 0 h` 4 x + 4 h + 5 + 4 x + 5j
4
= lim _ Now you can plug ini
h " 0 4 x + 4 h + 5 + 4 x + 5
4
=
4 x + $ 5 + 4 x + 5
4 0 +
4
=
2 4 x + 5
2
g x =
l ^ h
4 x + 5
2
k Use the parabola from problem 8, but make it a position function, s t = t 3 , where t is in hours
^ h
and s(t) is in miles.
total distance
Average velocity equals .
total time
a. What’s the average velocity from t = 4 to t = 5? The answer is 27 miles/hour.
^
s 5 - ^h s 4h
Average Velocity 4 to =
5
5 - 4
2
$
3 5 - $ 2
3 4
=
1
miles
= 27
hour
b. What’s the average velocity from t = 4 to t = 4.1? The answer is 24.3 miles/hour.
s ^ . 4 1 - ^h s 4h
Average Velocity 4 to = . 4 1 - 4
. 4 1
3 $ . 4 1 - $ 2
2
3 4
=
. 0 1
miles
= 24 .3
hour