Suppose f is a function from A to B where A and B are finite sets with |A| = |B|. Show that f is
one to one if and only if it is onto.
Solution
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Let f : A ---> B be function.
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Given |A = |B|
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Suppose that f is one to one function.
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i.e., f(x) = f(y) is B => x = y is A
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with respect to the function of f, every element in B has apre-image in A.
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f is onto.
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Suppose that f is onto function.
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Since each element is A has one and only one image in Band the number of elements in B
.
and the number of images of f are equal.
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It follows that distinct elements in A have distinct images inB.
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f is one to one.
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Hence f is one to one if and only if it isonto.
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Suppose f is differentiable on (a,b) and f is bounded on (a,b). .pdf
1. Suppose f is differentiable on (a,b) and |f'| is bounded on (a,b). Prove that f is uniformly
continuous on (a,b).
Solution
if f is differentaible on (a,b) , then mean value theorem states that , for some c
belongs to (a,b) , f'(c) = f(b) - f(a) / (b-a) hence , according to the definition of differentiation ,
uniformly continuous function should have a derivative , which f(x) has according to MVT and
also MVT states the definition at end points because the function is uniform on the interval (a,b)
for reference : he expression (f(b) - f(a)) / (b - a) gives the slope of the line joining the points (a,
f(a)) and (b, f(b)), which is a chord of the graph of f, while f'(x) gives the slope of the tangent to
the curve at the point (x, f(x)). Thus the Mean value theorem says that given any chord of a
smooth curve, we can find a point lying between the end-points of the curve such that the tangent
at that point is parallel to the chord. The following proof illustrates this idea. Define g(x) = f(x) -
rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is
true for g. We now want to choose r so that g satisfies the conditions of Rolle's theorem. Namely
g(a) = g(b) => f(b) -rb = f(a) - ra => r = f(b) - f(a) / (b-a) By Rolle's theorem, since g is
continuous and g(a) = g(b), there is some c in (a, b) for which , and it follows from the equality
g(x) = f(x) - rx that, f'(c) = g'(c) + r = 0 + r = r
![Suppose f is differentiable on (a,b) and |f'| is bounded on (a,b). Prove that f is uniformly
continuous on (a,b).
Solution
if f is differentaible on (a,b) , then mean value theorem states that , for some c
belongs to (a,b) , f'(c) = f(b) - f(a) / (b-a) hence , according to the definition of differentiation ,
uniformly continuous function should have a derivative , which f(x) has according to MVT and
also MVT states the definition at end points because the function is uniform on the interval (a,b)
for reference : he expression (f(b) - f(a)) / (b - a) gives the slope of the line joining the points (a,
f(a)) and (b, f(b)), which is a chord of the graph of f, while f'(x) gives the slope of the tangent to
the curve at the point (x, f(x)). Thus the Mean value theorem says that given any chord of a
smooth curve, we can find a point lying between the end-points of the curve such that the tangent
at that point is parallel to the chord. The following proof illustrates this idea. Define g(x) = f(x) -
rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is
true for g. We now want to choose r so that g satisfies the conditions of Rolle's theorem. Namely
g(a) = g(b) => f(b) -rb = f(a) - ra => r = f(b) - f(a) / (b-a) By Rolle's theorem, since g is
continuous and g(a) = g(b), there is some c in (a, b) for which , and it follows from the equality
g(x) = f(x) - rx that, f'(c) = g'(c) + r = 0 + r = r](https://image.slidesharecdn.com/supposefisdifferentiableonabandfisboundedonab-230328015440-79eaf4bc/85/Suppose-f-is-differentiable-on-a-b-and-f-is-bounded-on-a-b-pdf-1-320.jpg)
