suppose f is a continuous function on [-2,2] such that f(-2)=1 , f(2)= -1?

Question

Which of the properties below follow without further restriction on f by applying the Intermediate Value Theorem?
A) f^2 (c) is less than or equal to 1 for all c in (-2,2)
B) graph of f(x) +x crosses x-axis on (-2,2)
C) f(c)=0 for some c in (-2,2)

would b and c apply? what do they mean by f(-2)=1 f(2)= -1 so i know that i have to imagine a line from -2 to 2 there is 1 and -1?

Answer 1

As you have already said yourself: B and C apply.

First C.

Because f(-2) is positive, f(2) is negative and f is continuous, the graph of f(x) MUST cross x-axis somewhere between -2 and 2:
imagine drawing a line. The starting point is (-2, 1), the ending point is (2, -1). However curved (or straight) the line is, if you wan to connect the points (-2, 1) and (2, -1), you will have to cross the x-axis somewhere. In other words: f(c)=0 for some c in (-2,2).

Now B.

g(x) = f(x) + x
g(-2) = f(-2) + (-2) = 1 - 2 = -1
g(2) = f(2) + 2 = -1 + 2 = 1

This case is analogous to the case above, but now we start drawing the line below the x-axis (at the point (-2, -1)) and end above the x-axis (at the point (2, 1)). We have to cross the x-axis somewhere on the way.

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Hope this helps.

Answer 2

f(-2) is the value of y when x = -2
f(2) is the value of y when x = 2.
Since f(-2) is positive and f(2) is negative, f(x) must have crossed the x-axis at least once between x= -2 and x =2.
When f(x) crosses the x-axis at x = c, f(c) = y = 0.

So B and C are correct

Answer 3

A) option is wrong cuz u can imagine a continuos function whose value exceeds 1 for some "c" belonging to (-2,2)

by option B), u mean that
f(x)+x=0
ie: f(x) = -x
for a continuous function with given conditions, the line...y=-x, would surely intersect f(x), so v will have some value of x in the range (-2,2) to satisfy the condition : f(x) = -x
thus, B) IS CORRECT

C) option is obvious as the function is continuos and the values at end points are of opposite signs....OPTION C) IS ALSO CORRECT