So lets try to prove it,
So let's consider the function f(x) = x^2.
Since f(x) is a polynomial, then it is continuous on the interval (- infinity, + infinity).
Using the Intermediate Value Theorem,
it would be enough to show that at some point a f(x) is less than 2 and at some point b f(x) is greater than 2. For example, let a = 0 and b = 3.
Therefore, f(0) = 0, which is less than 2, and f(3) = 9, which is greater than 2. Applying IVT to f(x) = x^2 on the interval [0,3}.
Learn more about Intermediate Value Theorem on:
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Answer:
x= -3
Step-by-step explanation:
hello :
-2x+7=-6x-5
-2x+7+6x=-6x-5 +6x
4x+7=-5
4x+7-7=-5-7
4x= -12
x= -12/4
x= -3
Answer:
The input value is 3/4
Step-by-step explanation:
we know that
The input value that produces the same output value for the two linear functions, is the intersection point both graphs
we have
---> equation A
---> equation B
Equate equation A and equation B
solve for x
therefore
The input value is 3/4
I think the correct answer is 17
