Prove that a strictly decreasing function is always one-one
In real analysis, to prove that a strictly decreasing function is always one-to-one (injective), you can use the definition of a strictly decreasing function and the definition of a one-to-one function. A function \(f: A \to B\) is strictly decreasing if, for all \(x, y \in A\) with \(x < y\), we have \(f(x) > f(y)\). To prove that a strictly decreasing function
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Determine the points of discontinuity of the function f and the nature of discontinuity at each of those points:
Also check whether the function f is derivable at x = 1.
Apply Bonnet Mean Value Theorem for integrals to show that
Show that is an algebraic number.
Verify Bozano–Weierstrass Theorem for the following sets:
i) Set of non-negative integers.
ii) Interval [-1, ¥]
Check whether the limit exists or not?
Use Cauchy’s mean value theorem to prove that:
Verify the second mean value theorem for the function f(x) = x and
g(x) = cos in the interval