Explanation The given inequality is 1 2 ( x + 15 ) ≥ 5 − 2 x . Obtain the solution of the inequality 1 2 ( x + 15 ) ≥ 5 − 2 x using TI-83 graphing calculator as follows. Set the function y 1 = 1 2 ( x + 15 ) ≥ 5 − 2 x as shown below. Now press the WINDOW key and set x min = − 5 , x max = 5 , y min = − 1 , y max = 3 , x scl = 1 and y scl = 1 . Press the GRAPH key to obtain the graph of the inequality 1 2 ( x + 15 ) ≥ 5 − 2 x as shown below

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Chapter 17.2, Problem 31E

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The solution of the inequality 12(x+15)≥5−2x using a graphing calculator.

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Chapter 17.2, Problem 30E

Chapter 17.2, Problem 32E

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4. Let F RNR be a mapping. (i) x ЄRN ? (ii) : What does it mean to say that F is differentiable at a point [1 Mark] In Theorem 5.4 in the Lecture Notes we proved that if F is differentiable at a point x E RN then F is continuous at x. Proof. Let (n) CRN be a sequence such that xn → x ЄERN as n → ∞. We want to show that F(xn) F(x), which means F is continuous at x. Denote hnxn - x, so that ||hn|| 0. Thus we find ||F(xn) − F(x)|| = ||F(x + hn) − F(x)|| * ||DF (x)hn + R(hn) || (**) ||DF(x)hn||+||R(hn)||| → 0, because the linear mapping DF(x) is continuous and for all large nЄ N, (***) ||R(hn) || ||R(hn) || ≤ → 0. ||hn|| (a) Explain in details why ||hn|| → 0. [3 Marks] (b) Explain the steps labelled (*), (**), (***). [6 Marks]

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Chapter 17 Solutions

EBK BASIC TECHNICAL MATHEMATICS

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The solution of the inequality 1 2 ( x + 15 ) ≥ 5 − 2 x using a graphing calculator.

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The solution of the inequality 12(x+15)≥5−2x using a graphing calculator.

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