Question

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.

Answer 1

plzzz help I dont know what to do and i cant get help😭

Answer 2

like i tried everything i could but idk how to do it

Answer 3

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 <u>except</u> 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

Answer 4

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

Answer 5

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 ?