Answer:
f(x) = 2x^2 +3x - 1
f'(x) = 4x + 3
f''(x) = 4
f(0) = 2(0)^2 + 3(0) - 1 = -1 (valid)
f'(0) = 4(0) + 3 = 3 (valid)
f''(0) = 4 (valid)
=> f(x) is continous at x = 0
Answer:
3,-9,-6,-15,-21,-36,-57
Step-by-step explanation:
Answer:
i dont understand this question at all
Step-by-step explanation:
Answer:
Step-by-step explanation:
Find the absolute value vertex. In this case, the vertex for
y
=
|x
−
2
|
+
1
is (
2
,
1
)
.
To find the x coordinate of the vertex, set the inside of the absolute value
x
−
2 equal to 0
. In this case,
x
−
2
=
0.
x−2
=
0
Add 2 to both sides of the equation.
x
=
2
Replace the variable x with 2 in the expression.
y
=
|
(
2
)
−
2
|
+
1
Simplify
|
(
2
)
−
2
|
+
1
.
Simplify each term.
Subtract 2 from 2
y
=
|
0
|
+
1
The absolute value is the distance between a number and zero. The distance between 0 and 0 is 0
.
y
=
0
+
1
Add 0 and 1
.
y
=
1
The absolute value vertex is (
2
,
1
)
.
(
2
,
1
)
Answer:
x = 31
Step-by-step explanation:
Given:
f(x) =

g(x) = x + 2
We will first find g(2).
g(2) = 2 + 2 = 4
Next we will find f(g(2)).
f(g(2))= f(4) =
