Explanation Given: If v and w are any two vectors then | v ⋅ w | ≤ ‖ v ‖ ‖ w ‖ . Calculation: Part a: Let f ( x ) = ‖ x v + w ‖ 2 for x a scalar. Also, a = ‖ v ‖ 2 , b = 2 v ⋅ w , and c = ‖ w ‖ 2 . We have to prove that f ( x ) = a x 2 + b x + c . f ( x ) = ‖ x v + w ‖ 2 = ‖ x v ‖ 2 + ‖ w ‖ 2 + 2 x v ⋅ w = x 2 ‖ v ‖ 2 + ‖ w ‖ 2 + 2 x v ⋅ w It is given that, a = ‖ v ‖ 2 , b = 2 v ⋅ w , and c = ‖ w ‖ 2 . f ( x ) = x 2 ‖ v ‖ 2 + ‖ w ‖ 2 + 2 x v ⋅ w = x 2 ⋅ a + c + x ⋅ b = a x 2 + c + b x Hence, proved. Part b: We have to prove that b 2 − 4 a c ≤ 0 . It is given that, a = ‖ v ‖ 2 , b = 2 v ⋅ w , and c = ‖ w ‖ 2 ...

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Chapter 13.3, Problem 97E

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a. Let $f (x) = {‖ x v + w ‖}^{2}$ for $x$ a scalar. Prove that $f (x) = a x^{2} + b x + c$ , where $a = {‖ v ‖}^{2}, b = 2 v \cdot w, and c = {‖ w ‖}^{2}$ .

b. Conclude that $b^{2} - 4 a c \leq 0$ .

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Chapter 13.3, Problem 96E

Chapter 13.3, Problem 98E

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Chapter 13 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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a. Let f ( x ) = ‖ x v + w ‖ 2 for x a scalar. Prove that f ( x ) = a x 2 + b x + c , where a = ‖ v ‖ 2 , b = 2 v ⋅ w , and c = ‖ w ‖ 2 . b. Conclude that b 2 − 4 a c ≤ 0 .

Solution Summary: The author explains that if v and w are any two vectors then f (x) = a x 2 + b c.

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a. Let $f (x) = {‖ x v + w ‖}^{2}$ for $x$ a scalar. Prove that $f (x) = a x^{2} + b x + c$ , where $a = {‖ v ‖}^{2}, b = 2 v \cdot w, and c = {‖ w ‖}^{2}$ .

b. Conclude that $b^{2} - 4 a c \leq 0$ .

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Chapter 13 Solutions

a. Let f ( x ) = ‖ x v + w ‖ 2 for x a scalar. Prove that f ( x ) = a x 2 + b x + c , where a = ‖ v ‖ 2 , b = 2 v ⋅ w , and c = ‖ w ‖ 2 . b. Conclude that b 2 − 4 a c ≤ 0 .

Solution Summary: The author explains that if v and w are any two vectors then f (x) = a x 2 + b c.

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a. Let f(x)=‖xv+w‖2 for x a scalar. Prove that f(x)=ax2+bx+c, where a=‖v‖2,b=2v⋅w, and c=‖w‖2. b. Conclude that b2−4ac≤0.

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Chapter 13 Solutions

a. Let $f (x) = {‖ x v + w ‖}^{2}$ for $x$ a scalar. Prove that $f (x) = a x^{2} + b x + c$ , where $a = {‖ v ‖}^{2}, b = 2 v \cdot w, and c = {‖ w ‖}^{2}$ .

b. Conclude that $b^{2} - 4 a c \leq 0$ .