One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in x=0
x
=
0
.
Here's another way. Look at
π2−2arctanx−−√=2(π4−arctanx−−√)=2(arctan1−arctanx−−√).
π
2
−
2
arctan
x
=
2
(
π
4
−
arctan
x
)
=
2
(
arctan
1
−
arctan
x
)
.
Now remember the identity for the difference of two arctangents:
arctanu−arctanv=arctanu−v1+uv.
arctan
u
−
arctan
v
=
arctan
u
−
v
1
+
u
v
.
(This follows from the usual identity for the tangent of a sum.) The left side above becomes
2arctan1−x−−√1+x−−√.
2
arctan
1
−
x
1
+
x
.
The double-angle formula for the sine says sin(2u)=2sinucosu
sin
(
2
u
)
=
2
sin
u
cos
u
. Apply that:
sin(2arctan1−x−−√1+x−−√)=2sin(arctan1−x−−√1+x−−√)cos(arctan1−x−−√1+x−−√)
sin
(
2
arctan
1
−
x
1
+
x
)
=
2
sin
(
arctan
1
−
x
1
+
x
)
cos
(
arctan
1
−
x
1
+
x
)
Now remember that sin(arctanu)=u1+u2−−−−−√
sin
(
arctan
u
)
=
u
1
+
u
2
and cos(arctanu)=11+u2−−−−−√
cos
(
arctan
u
)
=
1
1
+
u
2
Then use algebra:
2⋅(1−x√1+x√)1+(1−x√1+x√)2−−−−−−−−−−√⋅11+(1−x√1+x√)2−−−−−−−−−−√=1−x1+x.
First term a1 = -2
and the common difference is + 3
Answer is choice D
Step-by-step explanation:
ax^2+bx+c=0
a=leading term
ok so if the leading term is positive then opens up and has a <u>min</u>
if leading term is negative then opens down and has a <u>max</u>
leading term is positive
1x^2+8x
it has a min
to complete the square, move c aside take 1/2 of b and square it
b=8
8/2=4
4^2=16
now add that to both sides
x^2+8x+16+6=0+16
factor perfect square
(x+4)^2+6=16
subtract 6
(x+4)^2=10
subtract 10
(x+4)^2-10=0
vertex aka min or max is (h,k) when ou have
y=a(x-h)+k
h=-4
k=-10
The answer might be D I don’t understand either
