Calculate the Taylor polynomials T₂(x) and T3(x) centered at x = 9 for f(x) = 1 T2(x) must be of the form where A equals: Bequals: C'equals: T3(x) must be of the form where D equals: E equals: Fequals: G equals: 8. and and A+ B(x-9) + C(x - 9)² D+E - 9) + F(x −9)² + G(x −9)³

The number of zeros of f(x) = 6x'– 7x +9x- 2 is , provided that each zero is counted according to its multiplicity.

Differentiate the following functions using product rule. 1. f (x) = (3x – 9) (x – 6) 2. f (æ) = (³ – 2x² + † – 6) (2æ² – 2æ + 4) 3 3 -

The difference quotient is defined as f(x+h)={(x) where h + 0. Find m for f(x) = x² – 9

Find all zeros of f(æ) = x' + 4x² + x + 4. Enter the zeros separated by commas.

Let f(x) = 3x^3 + 3x^2 + 3x+ 3. Write f(x) as a product of irreducible elements in:

5. If f(x)=√√x and g(x) = x²-1, find the functions (a) fog, (b) gof, (c) fof, (d) go g, and their domains.

Show that the equation x3 - 17x + c = 0 has at most one solution in the interval [-2, 2]. Let f(x) = x - 17x + c for x in [-2, 2]. If f has two real solutions a and b in [-2, 2], with a < b, then -Select--- polynomial f is continuous on [a, b] and differentiable on (a, b), -Select-- v. Since the ) implies that there is a number r in (a, b) such that f'(r) = 0. Now f'(r) = . Since r is in (a, b), which is contained in [-2, 2], we have Irl < 2, so r? < 4. It follows that 3r2 - 17 ?v 3 ·4 - 17 = -5 ?v 0. This contradicts f'(r) = 0, so the given equation cannot have two real solutions in [-2, 2]. Hence, it has at most one real solution in [-2, 2].

Q3. Using interpolation, approximate the following function by a polynomial of degree at most 3 using node points x; = i, i = -1,0, 1,2 : f(x) = 3x² – e*

2. Determine two non-equivalent polynomials f(x) and g(x), such that f(0) = g(0) and f(1) = g(1).