Answers and Explanations 233
Answers
201–300
To rationalize the numerator, consider the formula a
3
– b
3
= (a – b)(a
2
+ ab + b
2
). If you
let a = (2x + 2h + 1)
1/3
and b = (2x + 1)
1/3
, you have (a – b) in the numerator; that means
you can rationalize the numerator by multiplying by (a
2
+ ab + b
2
):
lim
lim
h
h
x
h
x
h
x
h
x
→
→
+ +
(
) − +
(
)
=
+ +
(
) − +
(
)




0
1 3
1 3
0
1 3
1 3
2
2 1
2
1
2
2 1
2
1
h h
x
h
x
h
x
x
x
h
2
2 1
2
2 1
2
1
2
1
2
2 1
2 3
1 3
1 3
2 3
2 3
+ +
(
) +
+ +
(
)
+
(
) +
+
(
)




+ +
(
) + 2 2 2 1 2 1
2
1
2
2 1
2
1
2
1 3
1 3
2 3
0
x
h
x
x
x
h
x
h
h
+ +
(
)
+
(
) +
+
(
)




=
+ +
(
)− +
(
)
→
lim
x x
h
x
h
x
x
h
h
x
h
+ +
(
) +
+ +
(
)
+
(
) +
+
(
)




=
+
→
2 1
2
2 1
2
1
2
1
2
2
2
2 3
1 3
1 3
2 3
0
lim
h h
x
h
x
x
x
h
h
+
(
) +
+ +
(
)
+
(
) +
+
(
)




=
+ +
(
)
→
1
2
2 1
2
1
2
1
2
2
2 1
2 3
1 3
1 3
2 3
0
lim
2 2 3
1 3
1 3
2 3
2 3
2
2 1
2
1
2
1
2
2
2 0 1
2
2 0
+
+ +
(
)
+
(
) +
+
(
)




=
+ ( )+
(
) + + (
x
h
x
x
x
x
) ) +
(
)
+
(
) +
+
(
)

 

 
=
+
(
) +
+
(
) +
+
(
)
1
2
1
2
1
2
2
1
2
1
2
1
1 3
1 3
2 3
2 3
2 3
2
x
x
x
x
x
3 3
2 3
2
3 2
1
=
+
(
)
x
291.
0
The tangent line at x = 3 is horizontal, so the slope is zero. Therefore, f '(3) = 0.
292.
–1
The slope of the tangent line at x = –1 is equal to –1, so f '(–1) = –1.
293.
1
The slope of the tangent line at x = –3 is equal to 1, so f '(–3) =1.
294.
3
The slope of the tangent line at any point on the graph of y = 3x + 4 is equal to 3,
so f '(–22π
3
) = 3.
295.
f '(1) < f '(–2) < f '(–3)
The tangent line at x = –3 has a positive slope, the slope of the tangent line at x = –2
is equal to zero, and the slope of the tangent line at x = 1 is negative. Therefore,
f '(1) < f '(–2) < f '(–3).