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x5y2 - 2 = 0

... makes sense

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∙ 13y ago
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Q: Is x 5y2 - 2 a function?
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X equals 5y2-20y plus 15?

x isn't in the calculation so it cant be solved.


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The "root" of a function is also called the "zero" of a function. This is where the function equals zero. The function y=4-x2 has roots at x=2 and x=-2 The function y=4-x2 has zeroes at x=2 and x=-2 Those are equivalent statements.


Is it true or false that the limit of a function f x at x equals 2 is always the value of the function at x equals 2 that is f 2?

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The function given is (f(x) = -x^2). The second derivative of a function, denoted as (f’'(x)), measures the concavity of the function. For the function (f(x) = -x^2), the first derivative (f’(x)) is (-2x). Taking the derivative of (f’(x)) gives us the second derivative (f’‘(x)), which is (-2). So, (f’'(x) = -2). This indicates that the function (f(x) = -x^2) is concave down for all (x), because the second derivative is negative.


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How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?


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