Use the functions f and g in C[-1, 1] to find (f, g), ||f||, |lg||, and d(f, g) for the inner product = [+ f(x)g(x)dx. (f. g) = f(x) = 5x, g(x) = ex (a) (f. g) (b) ||f|| (c) ||g|| (d) d(f, g)

Q: Verify that the factorization for f (x) = x3 + x2 + 1 over Z2

A: Given that, fx=x3+x2+1. Also ℤ2=0, 1 To verify the factorization for fx over ℤ2.

Q: Find f(x) and g(x) such that h(x) = (fo g)(x) and g(x) = Vx +9. h(x) = (Vx +9) * 4 f(x) = x

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Q: Define F : R → R and G : R → Z by the following formulas: F(x) = (x^2)/3 and G(x) = ⌊x⌋ for every x…

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Q: f(x) = -9 f'(x) = -(x²+9) (x²-9)2 f"(x) = 2x(x²+27) (x²–9)3 4. Given a) Provide a complete analysis…

A: a)

Q: Algebra Question

A: Solution: The objective is add the given number

Q: f(x) = sqrt 9-x, [-16,9]  If the Mean Value Theorem can be applied, find all values of c.

A: Consider the given function as fx=9-x and the interval is -16,9. Using a graphing utility, sketch…

Q: Determine F(x) if, for all real x and y, F(x)F(y) – F(xy) = x + y.   Hint:  What could F(1) be?

A: Consider the provided question, We have to find F(x).

Q: Use calculus to show that if f(x) = (x − a)²g(x), where both f(x) and g(x) are polynomials, then…

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Q: b) h(x) - 13 — 2х - х?). c) h(x) = V3 - x 5, d) h(x) = Vx³ + 4x2 + 1.

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Q: Exercise 5: If f(x) = x5 + 2x³ + 2x, check that f is injective and find (f-1)'(-5).

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Q: Define f (x) = Vx² + 1 and g(x) = 2 – x². Find and simplify (go f)(x).

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Q: Show that the product rule works for the case f (x) = x2, g (x) = x3.

A: Given that, fx=x2gx=x3 We have to show that the product rule works for the case of given functions.

Q: Let ƒ(x) = (2x + 3)(3x2). Use the product rule to find ƒ′(x).

A: Consider the formula for product rule. Product rule: fg'=f'⋅g+f⋅g'

Q: Let f(x)=x2, g(x)=2x+5, and h(x)=x2-1 a. h[f(-2)] b. f[g(x-1)]

A: Given that , fx=x2 gx=2x+5 hx=x2-1

Q: Write the composite function in the form Køtx)). [Identify the inner function u - g(x) and the outer…

A: We need to find out functions y=fu and u=gx We need to make a composition of the functions f and g…