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:
$10.80
Step-by-step explanation:
$2.40 + $2.80(3)
$2.40 + 8.40
(√10)/(√3)*(√3)/(<span>√3); multiplied by the conjugate which is = (</span><span>√30/3)</span>
Answer:
d / 2
Step-by-step explanation:
