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}.
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Answer:
steps below
Step-by-step explanation:
2ⁿ⁺¹ +2ⁿ⁻¹ = 2ⁿ⁻¹ (2² + 1) = 5*2ⁿ⁻¹
1)7; divide by 7 2)-3; divide by 2 3)8; divide by -4 4)54; multiply by 6 5)80; multiply by -10 6)-6; divide by -9 7)-5; divide by -12 8)-40; multiply by 20 9)-90; multiply by 10 10)20 11)615; plug in 205 for R and 3 in T. multiply by 3
The function f(x) takes on the value 2 when x = -7 and x = -3, and thus the corresponding points are (-7,2) and (-3,2).
The points where g(x) = 0 are called "roots" or "solutions," and is (3,0).
f(x) > g(x) on the approx. interval (-9, -3/4).
Answer:
<h2>560</h2>
Step-by-step explanation:
You must use a combination:
We have <em>n = 16</em>, <em>k = 3</em>.
Substitute:
