Answer:
a) (-2,2), (-1,-1), (1,-1), (3,7)
b) x=1.4
Answer:
Step-by-step explanation:
Correct option is
D
[1,(1+π)
2
]
f(x)=(1+sec
−1
(x))(1+cos
−1
(x))
Here the limiting component is cos
−1
(x), since the domain of cos
−1
(x) is [−1,1].
Therefore,
f(1)=(1+0)(1+0)
=1
f(−1)=(1+π)(1+π)
=(1+π)
2
Hence range of f(x)=[1,(1+π)
2
]
Solve by Factoring x^(2/3)-7x^(1/3)+10=0
x
2
3
−
7
x
1
3
+
10
=
0
x
2
3
-
7
x
1
3
+
10
=
0
Rewrite
x
2
3
x
2
3
as
(
x
1
3
)
2
(
x
1
3
)
2
.
(
x
1
3
)
2
−
7
x
1
3
+
10
=
0
(
x
1
3
)
2
-
7
x
1
3
+
10
=
0
Let
u
=
x
1
3
u
=
x
1
3
. Substitute
u
u
for all occurrences of
x
1
3
x
1
3
.
u
2
−
7
u
+
10
=
0
I believe the answer is $45.75.
I hope this helps.
If you would like an explanation of how I got this, please reply.
Answer:
you cannot answer this without a screen shot
Step-by-step explanation:
