Suppose f is a continuous function on [-2, 2) such that f(-2) = 1, f(2) = -1
Which of the properties below follow without further restriction on f by applying the Intermediate Value Theorem?
a. f(c) = 0 for some c in (-2, 2).
b. the graph of f(-x) + x crosses the x-axis on (-2, 2);.
c. f(c) < 1 for all c in (-2, 2).
1. b and c only.
2. a and b only.
3. b only.
4. a only
5. c only.
6. a and c only.
7. all of them.
8. none of them.

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.

goog may have it for you