What is the domain of the function f(x)= x + 1 / x^2 - 6x + 8?
1 answer:
Answer:
all real numbers except 4 and 2
Explanation:
The given function is:
f(x) =
Domain of the function is defined as the values of x that can be used to substitute in the function.
We know that any fraction cannot have a zero in its denominator. Because if this happens, then the fraction would be undefined.
This means that in the given function, we cannot substitute with any value for x that makes the denominator a zero.
Now, we will get the zeroes of the denominator as follows:
x² - 6x + 8 = 0
(x-4)(x-2) = 0
either x-4 = 0 ...........> x = 4
or x-2 = 0 .............> x = 2
Based on the above, we can substitute with any value for x excluding 4 and 2.
This means that the domain of the function is any real real number except 4 and 2
Hope this helps :)
all real numbers except 4 and 2
Explanation:
The given function is:
f(x) =
Domain of the function is defined as the values of x that can be used to substitute in the function.
We know that any fraction cannot have a zero in its denominator. Because if this happens, then the fraction would be undefined.
This means that in the given function, we cannot substitute with any value for x that makes the denominator a zero.
Now, we will get the zeroes of the denominator as follows:
x² - 6x + 8 = 0
(x-4)(x-2) = 0
either x-4 = 0 ...........> x = 4
or x-2 = 0 .............> x = 2
Based on the above, we can substitute with any value for x excluding 4 and 2.
This means that the domain of the function is any real real number except 4 and 2
Hope this helps :)
You might be interested in
Answer:
B) x = 7, y = 4
Step-by-step explanation:
you can find the value of 'y' first by connecting the bases with another altitute measuring 4 units
you now have an isosceles triangle where each leg is 4 which makes the hypotenuse equal to 4
to find 'x', it is the sum of 3 and 4
Using the fact that cos is 2π-periodic, we have
That is, for any
and integer
.
We get 2 solutions in the interval [0, 2π] for and
,