solutions4

# solutions4 - | x | → ∞ for large values of | x | the...

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CM221A ANALYSIS I Solutions to Sheet 4 1. We have tan x 1 + tan x = sin x sin x + cos x . If x < π/ 2 and x π/ 2 then sin x 1 and cos x 0. Applying the theorem about the limits of sums and quotients, we obtain lim x π/ 2 - 0 tan x 1 + tan x = 1 1 + 0 = 1. 2. The composition and sum of continuous functions are continuous functions. The quotient is also a continuous function provided that the denominator does not vanish. Since sin and cos are continuous functions, tan = sin cos , all polynomials are continuous and 1 + x + x 2 6 = 0, we see that tan( x + x 2 ) 1 + x + x 2 is continuous everywhere with the exception of points x satisfying cos( x + x 2 ) = 0. In the interval [0 , 1] there is only one such point, namely, the solution to the equation x + x 2 = π/ 2. 3. The ﬁrst part is proved by evaluating the left hand side at the end points and applying the intermediate value theorem. The precise formula for the solution is obtained by substituting s = cos x . 4. Since (1 + x 2 ) - 1 0 as
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Unformatted text preview: | x | → ∞ , for large values of | x | the function sin x-(1+ x 2 )-1 is positive near the points π/ 2+2 nπ and is negative near the points-π/ 2 + 2 nπ . By the intermediate value theorem, there is a solution in each interval [-π/ 2 + 2 nπ,π/ 2 + 2 nπ ], where n = 1 , 2 ,... 5. continuous and unbounded on [0 , 1); YES e.g. f ( x ) = 1 / (1-x ). 6. continuous and unbounded on [0 , 1]; NO, the range of a continuous function on a closed bounded interval is a closed bounded interval. 7. continuous on [0 , 1] and { f ( x ) : x ∈ [0 , 1] } = (0 , 1); NO, for the same reason 8. continuous on (0 , 1) and { f ( x ) : x ∈ (0 , 1) } = [0 , 1]; YES, e.g. f ( x ) = sin(4 πx ). 9. continuous on (0 , 1) and { f ( x ) : x ∈ (0 , 1) } = [0 , 1] ∪ [3 , 4]. NO, by the intermediate value theorem. 1...
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