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.

explanation:

step-by-step explanation:

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1