f(x)=x + 2/x^2 -9 and g(x)=11/x^2+3x
(a) find f(x) + g(x)
(b) list all of the excluded values
(c) classify each type of discontinuity
To receive credit, this must be done by Algebraic methods, not graphing
Understand and Think : what is being asked in the problem and what does that mean?
same equation as first one
(a) find f(x) + g(x)
(b) list all of the excluded values
(c) classify each type of discontinuity
To receive credit, this must be done by Algebraic methods, not graphing
same equation
(a) find f(x) + g(x)
(b) list all of the excluded values
(c) classify each type of discontinuity
To receive credit, this must be done by Algebraic methods, not graphing
Determine the points of discontinuity for the following rational function.
y= x+3/x^2 +8x+15
What are the points of discontinuity? Select all that apply.
A. x=0
B. y= - 3
C. x= - 3
D. x= - 5
E. y =0
F. y= - 5
G. There are no points of discontinuity.
why do you post the same problem THREE TIMES?
f(x)=x + 2/x^2 -9 and g(x)=11/x^2+3x
so, just use these values to find (I'm assuming the usual carelessness with parentheses here, as well as the evidence of rational functions)
f(x) + g(x) = (x+2)/(x^2-9) + 11/(x^2+3x)
= (x+2) / (x+3)(x-3) + 11 / x(x+3)
Now the common denominator is x(x+3)(x-3) so we now have
[(x+2)(x) + 11(x-3)] / x(x^2-9)
= (x^2+13x-33) / x(x^2-9)
The excluded values are those where the denominator is zero, since division by zero is undefined.
Those would be x = -3, 0, 3
Since the numerator is not zero at any of those values, they are all vertical asymptotes (infinite discontinuities)
-------------------------------------------------------------
y= x+3/x^2 +8x+15 = (x+3) / (x+3)(x+5)
Here, there are two discontinuities, at x = -3, -5
Note that there is a discontinuity at x = -3, but the simplified function is
y = 1/(x+5)
everywhere except at x = -3, since 0/0 is undefined. That means that there is a hole (removable discontinuity) at x = -3, which could be removed by defining y(-3) = 1/2
There is a vertical asymptote at x = -5 since y = 1/0 which is undefined and cannot be removed
y= x+3/x^2 +8x+15 = (x+3) / [ (x+3)(x+5) ] = 1/(x+5)
trouble if x = -5
but what if before we simplified x=-3
we get 0 / 0 ?
f(x)=x + 2/x^2 -9 and g(x)=11/x^2+3x
(a) find f(x) + g(x)
(b) list all of the excluded values
(c) classify each type of discontinuity
================================
I guess maybe you mean:
f(x)= (x + 2) / (x^2 -9 ) and g(x)=11/(x^2+3x)
(a) find f(x) + g(x)
(b) list all of the excluded values
(c) classify each type of discontinuity
====================================
f(x)= (x + 2) / [(x-3)(x+3]and g(x)=11/ x(x+3)
f(x)= x(x + 2) / x [(x-3)(x+3]and g(x)=11(x-3)/ [x(x+3)(x-3)]
f(x)= (x^2+2x) / x [(x-3)(x+3] and g(x)=(11x-33) / [x(x+3)(x-3)]
so f(x) + g(x)
= (x^2 +13 x -33) / [x(x+3)(x-3)]
That obviously explodes if x = 0 or x=3 or x = -3
plzzz help I dont know what to do and i cant get help😭
like i tried everything i could but idk how to do it
In order to solve these algebraic problems, we need to evaluate the given expressions and simplify them using algebraic methods.
Let's start with the first question:
(a) To find f(x) + g(x), we need to add the two given functions together:
f(x) = x + 2/x^2 - 9
g(x) = 11/x^2 + 3x
To add these functions, we need a common denominator for the fractions. The common denominator here is x^2:
f(x) + g(x) = (x + 2/x^2 - 9) + (11/x^2 + 3x)
= (x + 11)/x^2 + (2 - 9 + 3x)/x^2
= (x + 11 + 2 - 9 + 3x)/x^2
= (4x + 4)/x^2
= 4(x + 1)/x^2
Therefore, f(x) + g(x) = 4(x + 1)/x^2
(b) To find the excluded values, we need to identify the values of x that would make the denominators of the fractions equal to zero. In this case, the denominators are x^2 and x^2 + 3x.
The excluded values are the values of x that make the denominators zero:
For x^2, there are no excluded values, because a non-zero number squared is always positive.
For x^2 + 3x, we can factor out x to find the values of x that make it zero:
x(x + 3) = 0
x = 0 or x = -3
So, the excluded values are x = 0 and x = -3.
(c) To classify the types of discontinuity, we need to analyze the behavior of the function as x approaches the excluded values.
For x = 0, the function f(x) + g(x) has a removable discontinuity because the expression 4(x + 1)/x^2 can be simplified and defined at x = 0 by canceling out the common factors.
For x = -3, the function f(x) + g(x) has a vertical asymptote because the expression 4(x + 1)/x^2 approaches infinity as x approaches -3 from both sides.
Therefore, the types of discontinuity are: removable discontinuity at x = 0 and vertical asymptote at x = -3.
For the second question:
To determine the points of discontinuity for the rational function y = (x + 3)/(x^2 + 8x + 15), we need to identify the values of x that make the denominator equal to zero.
Factorizing the denominator:
x^2 + 8x + 15 = (x + 3)(x + 5)
The denominator is zero when x + 3 = 0 or x + 5 = 0.
So, the points of discontinuity are x = -3 and x = -5.
Therefore, the correct answers for the multiple-choice question are:
D. x = -3
F. x = -5