Question

(1 point) If p(x) and q(x) are arbitrary polynomials of degree at most 2, then the mapping <p,q> = p(-3)q(-3) + p(0)q(0) + p(3)q(3) defines an inner product in P_3. Use this inner product to find <p,q>, ||p||, ||q||, and the angle ? between p(x) and q(x) for p(x) = 4x^2 + 6 and q(x) = 4x^2 - 3x. <p,q> = ||p|| = ||q|| = ? = radians

          (1 point) If p(x) and q(x) are arbitrary polynomials of degree at most 2, then the mapping <p,q> = p(-3)q(-3) + p(0)q(0) + p(3)q(3) defines an inner product in P_3. Use this inner product to find <p,q>, ||p||, ||q||, and the angle ? between p(x) and q(x) for p(x) = 4x^2 + 6 and q(x) = 4x^2 - 3x.

<p,q> =
||p|| =
||q|| =
? = radians

(1 point) If p(x) and q(x) are arbitrary polynomials of degree at most 2, then the mapping <p,q> = p(-3)q(-3) + p(0)q(0) + p(3)q(3) defines an inner product in P3. Use this inner product to find <p,q>, ||p||, ||q||, and the angle ? between p(x) and q(x) for p(x) = 4x^2 + 6 and q(x) = 4x^2 - 3x.

<p,q> =
||p|| =
||q|| =
? = radians

Added by Kevin I.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

12/16/2022

Step 1

First, we need to find the inner product <p,q>: <p,q> = p(-3)q(-3) + p(0)q(0) + p(3)q(3) Given p(x) = 4x^2 + 6 and q(x) = 4x^2 - 3x, we can find the values of p(-3), p(0), p(3), q(-3), q(0), and q(3): p(-3) = 4(-3)^2 + 6 = 42 p(0) = 4(0)^2 + 6 = 6 p(3) = 4(3)^2 Show more…

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If p(x) and q(x) are arbitrary polynomials of degree at most 2, then the mapping <p,q> = p(-3)q(-3) + p(0)q(0) + p(3)q(3) defines an inner product in P3. Use this inner product to find <p,q>, ||p||, ||q||, and the angle Î¸ between p(x) and q(x) for p(x) = 4x^2 + 6 and q(x) = 4x^2 - 3x. <p,q> = ||p|| ||q|| Î¸ radians