Question

If f(x) and g(x) are arbitrary polynomials of degree at most 2, then the mapping ?f, g? = f(-2)g(-2) + f(0)g(0) + f(3)g(3) defines an inner product in P2. Use this inner product to find ?f, g?, ||f||, ||g||, and the angle ?f,g between f(x) and g(x) for f(x) = 2x^2 + 3x + 2 and g(x) = 2x^2 - 2x - 7. ?f, g? = ||f|| = ||g|| = ?f,g =

          If f(x) and g(x) are arbitrary polynomials of degree at most 2, then the mapping

?f, g? = f(-2)g(-2) + f(0)g(0) + f(3)g(3)

defines an inner product in P2. Use this inner product to find ?f, g?, ||f||, ||g||, and the angle ?f,g between f(x) and g(x) for

f(x) = 2x^2 + 3x + 2 and g(x) = 2x^2 - 2x - 7.

?f, g? =
||f|| =
||g|| =
?f,g =

If f(x) and g(x) are arbitrary polynomials of degree at most 2, then the mapping

?f, g? = f(-2)g(-2) + f(0)g(0) + f(3)g(3)

defines an inner product in P2. Use this inner product to find ?f, g?, ||f||, ||g||, and the angle ?f,g between f(x) and g(x) for

f(x) = 2x^2 + 3x + 2 and g(x) = 2x^2 - 2x - 7.

?f, g? =
||f|| =
||g|| =
?f,g =

Added by Kelly F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

05/23/2022

Step 1

** Given that (f,g) = f(-2)g(-2) + f(0)g(0) + f(3)g(3), we can substitute f(z) = 2x^2 + 3x + 2 and g(z) = 2x^2 - 2x into the formula: (f,g) = (2(-2)^2 + 3(-2) + 2)(2(-2)^2 - 2(-2)) + (2(0)^2 + 3(0) + 2)(2(0)^2 - 2(0)) + (2(3)^2 + 3(3) + 2)(2(3)^2 - 2(3)) (f,g) = Show more…

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If f(x) and g(x) are arbitrary polynomials of degree at most 2, then the mapping ✈f, g✉ = f(-2)g(-2) + f(0)g(0) + f(3)g(3) defines an inner product in P2. Use this inner product to find ✈f, g✉, ||f||, ||g||, and the angle αf,g between f(x) and g(x) for f(x) = 2x^2 + 3x + 2 and g(x) = 2x^2 - 2x - 7.