AP test Problem?

Question

Given the two functions f and h such that f(x)=x^3 -4x^2+x+6 and h(x)= f(X)/(x-3) , for x is not equal 3 and P, for x=3
A) Find all zeros of the function f
B) Find the value of p so that the function h is continuous at x=3.

This is a tough problem, I couldn't find the zeros, I did many way to simplify it to find the zeros but nothing worked so anyone know how to do it, plz help me
C) Using the value of p found in (B), determine whether h is an even function.

Answer 1

Given
f(x) = x³ - 4x² + x + 6
h(x) = f(x) / (x - 3), x ≠ 3
h(x) = P, x = 3

a) Find the zeros of the function f(x).

Factor f(x) and set equal to zero. Since the lead coefficient is one, the only possible rational roots are factors of the constant 6. Namely, ±1, ±2, ±3, ±6. If none of those work, there are no rational roots. Use synthetic division to try the various possibilities.

f(x) = (x + 1)(x² - 5x + 6) = (x + 1)(x - 2)(x - 3) = 0
x = -1, 2, 3
______________

B) Find the value of p so that the function h is continuous at
x = 3.

h(x) = f(x) / (x - 3) = (x + 1)(x - 2)(x - 3) / (x - 3)
h(x) = (x + 1)(x - 2)

h(3) = (3 + 1)(3 - 2) = 4*1 = 4 = P
__________

C) Using the value of p found in (B), determine whether h is an even function.

For an even function f(x) = f(-x).

h(x) = (x + 1)(x - 2) = x² - x - 2

h(3) = (3 + 1)(3 - 2) = 4*1 = 4
h(-3) = (-3 + 1)(-3 - 2) = (-2)(-5) = 10

h(3) ≠ h(-3) therefore h is not an even function.

Answer 2

f(x)=x^3 -4x^2+x+6
h(x)= f(X)/(x-3) for x not equal to 3
P for x = 3
Find:
a) all zeros of the function f
b) value of p so that function h is continuous at x = 3

Solution:
a) f(x)=x^3 -4x^2+x+6
f(x) = 0 = x^3 - 4x^2 + x + 6
Using synthetic division:

1 - 4 + 1+ 6 | 2
___2_- 4 - 6__
1 -2 - 3 0
Hence, the factors are
(x^2 - 2x - 3)(x - 2) = 0
(x - 3)(x + 1)(x - 2) = 0
x - 3 = 0,
x = 3 ANS.

x + 1 = 0
x = - 1 ANS

x - 2 = 0
x = 2 ANS.

b) h(x)= f(X)/(x-3)
h(x) = (x^3 -4x^2+x+6)/(x-3)
for the function h to be continuous at x = 3, the following must be met:
1) h(3) must exist
2) Lim of h as x approaches 3 exists
3) h(3) = Lim of h as x approaches 3
h(x) = [(x - 3)(x + 1)(x - 2)]/(x-3)
h(x) = (x+1)(x-2)
h(3) = (3 + 1)(3 - 2)
h(3) = (4)(1)
h(3) = 4
Lim[(x+1)(x-2)] = (3 + 1)(3 - 2) = (4)(1) = 4
as x---->3


teddy boy

Answer 3

1. note that the 0s of h are the 0s of f except that 3 is not a 0 of h unless p = 0.
By factoring f(x) = (x+1)(x-2)(x-3), so the 0s of f are -1, 2, 3.

lim(x->3) h(x) = lim(x->3) (x+1)(x-2)(x-3)/(x-3)
= lim(x->3) (x+1)(x-2) = 4, so let p=4

h is not an even function since h(-1) = 0, but h(1) = -2. Note that no function can be odd or even if there exists a root whose negative is not a root.

Answer 4

There are no rational zero as factors of the constant term 6 (ie 1,2,3,6,-1,-2,-3,-6) are not roots. So solve by an iterative process eg Newton ralphsons for the first root after finding a graphical or try and error estimate.
I get roots1.572, 3,514,and -1.086

Answer 5

once you call, you get asked on your AP type or social protection type, date of delivery, then credit card type. If it keeps asserting blunders, they probaly did no longer value you greater.