Function and Domain?

Question

The function f(x) = x^3 + 5x^2 + 4x find f(2) and when f(x)=0


find the domain of this function
f(x)= x^2-4
---------
x^2-1


I have no idea how to do this. If someone could help me out that would be awesome.

Answer 1

For the First problem, substitute 2 in for x. Then set the equation equal to 0 to solve for the second.

For the domain, you cannot have a zero in the denominator, so you need to find what makes the denominator zero. Set the denominator equal to zero and then solve.

0=x2-1
0=(x+1)(x-1)
x=1 and -1

so x cannot be 1 and -1. Your domain for the function is
(-infinity, -1) U (-1, 1) U (1, infinity)

Answer 2

f(2) = 2^3 + 5*2^2+4*2 = 8+ 20 +8 =36
f(x) = x(x^2 +5x+4) =x(x+4)(x+1) so f(x) = 0 when x = 0, -1 and -4

The denominator will be 0 when x = +/- 1.
So the function will be discontinuous at thes two x values as you cannot divide by zero. So the domain is all real values of x except x = -1 and x = 1.

Answer 3

F(2) mean you just substitute 2 for 'x'
there f(2) = 2^3 + 5.2^2 + 4.2 = 8 + 20 + 8 = 36.

Domain means the values of 'x' which the function can take.

the denominator is x^2 -1 = (x+1)(x-1)
As you can see, if x = 1 or -1, the function will attain a 0 in the denominator. Which means that function is _not defined_ at that point.

PS. i didnt understand what you meant by f(x) =0, so please clarify!

So the domain of the function would be = R - {1,-1}

Which means that it can take all the values in Real space (R) except for 1 and -1.

If this helps you please vote!!

Answer 4

(1) To find f(2) simply insert 2 for each x.

f(2) = 2^3 + 5(2)^2 + 4(2) = 8 + 20 +8 = 36.

(1a) first, factor your function.

f(x) = x(x^2 + 5x + 4) = x(x+1)(x+4)

For that to equal zero, x must be 0, -1, or -4.

(2) I'm not sure what your function is... if it's x^2-4/(x^2-1)...

The domain is all numbers where x^2-1<>0 since you cannot divide by zero.

By factoring, that means (x-1)(x+1)<>0 or x cannot be 1 or -1.

Answer 5

f(2) =
(2)^3 + 5(2)^2 + 4(2) =
8 + 5(4) + 8 = 36.

x^3 + 5x^2 + 4x = 0
x(x^2 + 5x + 4) = 0
x(x + 4)(x + 1) = 0
The three roots or solutions are:
x = 0.

x + 4 = 0
x = -4.

x + 1 = 0
x = -1.

The domain of f(x) = (x^2 - 4) / (x^2 - 1) is all Real numbers except x = ±1. [Since, division by zero is undefined, x^2 - 1 ≠ 0.]

Answer 6

properly area of the definition of a function is that for each selection interior the area, there is in basic terms one million ensuing fee interior the form (The area is the enter, the x values, to that end the numbers interior the area are -2, -one million, 0, one million, 2. whilst Written interior the type (x,y), x represents values interior the area, and y represents values interior the form.) As you will see out of your set of numbers, it is actual, so the courting between the numbers is a function.

Answer 7

for the 1st one, jus put the 2 in place for all the x's