Answer:
2. a and b only.
Step-by-step explanation:
We can check all of the given conditions to see which is true and which false.
a. f(c)=0 for some c in (-2,2).
According to the intermediate value theorem this must be true, since the extreme values of the function are f(-2)=1 and f(2)=-1, so according to the theorem, there must be one x-value for which f(x)=0 (middle value between the extreme values) if the function is continuous.
b. the graph of f(-x)+x crosses the x-axis on (-2,2)
Let's test this condition, we will substitute x for the given values on the interval so we get:
f(-(-2))+(-2)
f(2)-2
-1-1=-3 lower limit
f(-2)+2
1+2=3 higher limit
according to these results, the graph must cross the x-axis at some point so the graph can move from f(x)=-3 to f(x)=3, so this must be true.
c. f(c)<1 for all c in (-2,2)
even though this might be true for some x-values of of the interval, there are some other points where this might not be the case. You can find one of those situations when finding f(-2)=1, which is a positive value of f(c), so this must be false.
The final answer is then 2. a and b only.
Answer:
id.k but i need points sorry
Step-by-step explanation:
Answer:
x y
-2 5
-1 . 1
3 . 1
-1 -2
Step-by-step explanation:
in 1 just put in
x y
-2 5
-1 . 1
3 . 1
-1 -2
Answer:
x = 5
Step-by-step explanation:
vertical angles are congruent , so
∠ A = ∠ B , that is
5x - 1 = 4x + 4 ( subtract 4x from both sides )
x - 1 = 4 ( add 1 to both sides )
x = 5
Answer:
sin²x = (1 - cos2x)/2 ⇒ proved down
Step-by-step explanation:
∵ sin²x = (sinx)(sinx) ⇒ add and subtract (cosx)(cosx)
(sinx)(sinx) + (cosx)(cosx) - (cosx)(cosx)
∵ (cosx)(cosx) - (sinx)(sinx) = cos(x + x) = cos2x
∴ - cos2x + cos²x = -cos2x + (1 - sin²x)
∴ 1 - cos2x - sin²x = (1 - cos2x)/2 ⇒ equality of the two sides
∴ (1 - cos2x) - 1/2(1 - cos2x) = sin²x
∴ 1/2(1 - cos2x) = sin²x
∴ sin²x = (1 - cos2x)/2
