Find the zeros of the function: f(x) = [(1/ x+1) / (1/ x-2)]?

Answer 1

(1 / x+1) / (1 / x-2) = (x-2) / (x+1)
When x=-1, there is no answer.
That equals zero only when x=2.

Answer 2

If you are allowed to manipulate it symbolically first, you get
(x - 2) / (x + 1). The single zero of this function is at x = 2.

However, if it must be evaluated as-is, there are no zeroes, because to compute f(2) you must divide 1 by zero before dividing that into 1/3; therefore, it does not exist at 2.

Now, the value x = -1 is an asymptote. The function falls rapidly when x approaches -1 from the left, and becomes arbitrarily large (in the negative direction), and it becomes arbitrarily large when approaching from the right. The limits are -inf (infinity) from the left and +inf from the right.

Answer 3

f(x) = [ (1/ (x + 1) ] / [1 / (x - 2) ]

This is a complex fraction (fractions within fractions). To turn this into a simple fraction you have to multiply top and bottom by
(x + 1)(x - 2). This leaves us with

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

Now we can find the zeroes easier. The zeros of a function are defined for when f(x) = 0, so

0 = (x - 2) / (x + 1)

The only thing that makes this equal to 0 is when the numerator is equal to 0. That is

x - 2 = 0

Therefore, x = 2

Answer 4

f(0) = f(0 + 0) = f(0) + f(0) + 0^2*0 +0*0^2 = 2f(0), so f(0) =0. f(x + y) = f(x) + f(y) + x^2y + xy^2, so subtracting f(y) and dividing by way of x you have, (f(x+y) - f(y))/x = f(x)/x + xy + y^2. Now purely take the decrease of the two factors as x-->0, then {by way of the definition of spinoff} f '(y) = lim (f(x + y) - f(y))/x = lim {f(x)/x + xy + y^2} = a million + y^2. so which you have f '(x) = a million + x^2, and f '(0) = a million. in reality f(x) = x + x^3/3, and you're able to be able to actual verify applying the binomial theorem that f(x + y) = x + y + (x + y)^3/3 = x + x^3/3 + y + y^3/3 + xy^2 + yx^2 = f(x) + f(y) + xy^2 + yx^2. you will see that that it is frequently genuine for any polynomial f, back applying the binomial theorem, that f(x + y) = f(x) + f(y) + different stuff - in reality it fairly is much extra ordinary. the factor is you may help different words, the only time you have f(x + y) = f(x) + f(y) is while f = mx.

Answer 5

f(x) = (x - 2 ) / (x + 1) = 0 when x = 2

Answer 6

from the domain of f(x), x!=-1, x!=2
and this function has no zero point!
Because the 1/(x+1)!=0

Answer 7

f(x) = [(1/( x+1) / (1/( x-2)]
f(x) = (x - 2)/(x - 1) = 0
x = 2 is a zero of the function.

Answer 8

-1 and 2. because you cant divide by zero

Answer 9

(x-2)/(x+1) is the equivalent of your equation, which implies that its zero is 2.

Answer 10

this is called poles not zeros , because they made the equation undefined(infinity),
and simply they are:
rearranging the equation we will have
x-2/x+1
so you will have one zero at :
x=2
and one pole at:
x=-1