Calculus!!!?

Question

Given the function f defined by f(x)=2x-2 / x^2 + x-2

a. for what values of x is f(x) discontinuous?

b. at each point of discontinuity found in part (a), determine whether f(x) has a limit and, if so, give the value of the limit.

c. write an equation for each vertical and horizontal asymptote to the graph of f. justify your answer

d. a rational function g(x) = a/ b+x is such that g(x)=f(x) wherever f is defined. Find the values of a and b.

Answer 1

Given the function f defined by f(x)=2x-2 / x^2 + x-2

assuming you mean: f(x)= (2x-2) / (x^2 + x-2)
(without the parenthesis it is a completely different matter...)

a. for what values of x is f(x) discontinuous?
x^2 +x -2 = 0
(x+2)(x-1) = 0
so the function is discontinuous at x=-2 and x=1.
actually it is NOT even defined for those points.
other than that, the function is continuous.

b. at each point of discontinuity found in part (a), determine whether f(x) has a limit and, if so, give the value of the limit.

lim x->-2 (2x-2) / (x^2 + x-2) , observe that
2(-2) -2 =-6, so the limit DOES NOT exist.

lim x->1 (2x-2) / (x^2 + x-2) = lim x->1 2(x-1) / (x+2)(x-1)
= lim x->1 2/ (x+2) = 2/(1+2) = 2/3

c. write an equation for each vertical and horizontal asymptote to the graph of f. justify your answer

vertical asymptotes are lines of the form x=a where a is a point where f(x) is not defined and the limit does not exist!
so there is 1 vertical asymptote:
x=-2

horizontal asymptotes, you need to find
lim x-> +- infinity (2x-2) / (x^2 + x-2) =0
so the horizontal asymptote is
y=0

d. a rational function g(x) = a/ b+x is such that g(x)=f(x) wherever f is defined. Find the values of a and b.

(2x-2) / (x^2 + x-2) = 2/ (x+2)
so a=2 and b=2
..............................

Answer 2

a. Note that for (2x - 2) / (x^2 + x - 2), you can factor the top and the bottom. What we're interested in this case is what makes the denominator zero (for the whole function to be undefined).

So we factor the denominator: x^2 + x - 2 = (x + 2) (x - 1)

Therefore the function is undefined at x = -2 and x = 1. As a result, the function is discontinuous at x = -2 and x = 1.

b. Calculate lim ( x --> -2, (2x-2)/[(x+2)(x-1)] )
= lim (x -> -2, 2(x-1)/[(x+2)(x-1)])
= lim (x -> -2, 2/(x+2))

Since plugging in x = -2 for 2/(x+2) yields 2/0, the limit does not exist. What you look for in knowing whether a limit may exist is for the form 0/0.

We similarly reduce the function for the other limit.

lim (x -> 1, 2/(x+2)) = 2/(1+2) = 2/3

c. To find the vertical asymptote, you have to use the value where the limit does not exist. In this case, x = -2 is the vertical asymptote. (x = 1 is not).

To obtain the horizontal asymptote, you must take the limit as x -> infinity of the function and the limit as x -> -infinity.

First, divide all terms by the highest degree (in this case, it's x^2). So we have

lim (x --> infinity, (2/x - 2/x^2)/(1 + 1/x - 2/x^2) ) = (0 - 0)/(1 + 0 - 0) = 0.

Therefore, the horizontal asymptote is y = 0.

lim (x -> -infinity) yields the same result.

If g(x) = f(x), then

a/(b+x) = (2x-2)/(x^2 + x - 2)
a/(b+x) = 2(x-1)/[(x-1)(x+2)]

a(x-1)(x+2) = 2(x-1)(b+x)

I can't seem to get anywhere with this, so I'll leave it to someone else to answer.

Answer 3

i could say shape is element of engineering meaning that a reliable carry close on math is significant, pre-cal is is like algebra to the subsequent point with the Pi chart and understanding approximately radians and attitude measures (which seems significant) yet Calculus is the learn of derivatives, (the slope of a line at a undeniable factor) form of pointless no count what container of workd your going to, in case you inquire from me.