almost time for the calculus final, please help me solve!?

Question

A function f(x) is said to have a removable discontinuity at x=a if:
1. f is either not defined or not continuous at x=a
2. f(a) could either be defined or redefined so that the new function IS continuous at x=a

A.
Let f(x) = (2x^2 + 4x-16)/(x-2)
Show that f(x) has removable discontinuity at x=2 and determine what value for f(2) would make f(x) continuous at x=2.
Define f(2)..........???

B.
Let f(x) = (7/x) +(-6x+7)/x(x-1)
if x does not = 0
and 7 if x=0
show f(x) has removable discontinuity at x=0 and determine what value for f(0) would make f(x) continuous at x=0.
Define f(0)..........???

C.
Let f(x) = x^2+10x+27 if x<-5
1 if x=-5
-x^2-10x-23 if x>-5
Show that f(x) has a removable discontinuity at x=-5 and determine what value for f(-5)would make f(x) continuos at x=-5.
Must redefine f(-5)=????

Answer 1

## First question:

Choice 2 is correct. Choice 1 is more general - it describes all discontinuities. Choice 2 is specific to removable ones (you could "fill in the gap" so to speak).

## A.

You need to factor the numerator. You get:

f(x) = 2(x-2)(x+4) / (x-2)

Because f is not defined at x=2, it's okay to cancel. The function f can now be written as:

f(x) = 2(x+4) for x≠2.

This is a removable discontinuity. To remove it, simply redefine your function as:

f(x) = 2(x+4) for all numbers x

This means f(2) should equal 2(2+4) = 12

## B.

You need to combine the two fractions into one. The resulting expression is:

f(x) = x / x(x-1) for x≠1 , x≠0

Which we can reduce to:

f(x) = 1 / (x-1) for x≠1 , x≠0

And then remove the discontinuity:

f(x) = 1 / (x-1) for x≠1

Note now f(0) = -1

##C. Here, the function is not defined correctly at x=-5. To be continuous, we must give it a value so that both expressions for x agree, on the left of -5 and on the right. So plug -5 into both expressions:

(-5)² + 10(-5) + 27 = 2
-(-5)² -10(-5) -23 = 2

Because each "half" of the function agrees, we know this discontinuity is removable. Notice that the value 1 at x=-5 isn't important. To redefine it, we simply need to include -5 in the domain for one of the other pieces. Either of the following would be correct:

f(x) =
x² + 10x + 27 if x ≤ -5
-x^2 - 10x - 23 if x > -5

f(x) =
x² + 10x + 27 if x < -5
-x^2 - 10x - 23 if x ≥ -5

f(x) =
x² + 10x + 27 if x ≤ -5
-x^2 - 10x - 23 if x > -5
2 if x=-5

although the last one is a bit redundant.

Answer 2

A. f(x) = 2*(X - 2)*(x +4)/(x - 2) , if x NE 2. If x NE 2, we could cancel the factor x - 2 from numerator and denominator, and that would not change the value of f WHEN x NE 2. Thus if we redefine f(2) = 12, then f(x) = 2*(x + 4) for all x, and f is continuous at 2.

B. f(x) = 7/x + (-6x + 7)/(x*(x - 1)) if x NE 0, and f(0) = 7. Notice that f(x) = (7*(x-1) - 6x + 7)/(x*(x - 1))
= (7x - 7 - 6x + 7)/(x*(x - 1)) = x/(x*(x - 1)) when x NE 0. But when x NE 0, we can cancel the x factors and get
f(x) = 1/(x-1). Thus, if we redefine f(0) = -1, we then have f continuous at 0.

C. You should get the point by now.

Answer 3

dT/dt = -ok(T-40) says that the fee of replace of the temperature is proportional to its distinction from 40 deg. additionally, ok is constructive, so the temperature is changing to head in the direction of 40 deg. i.e. the better above 40 deg the temperature is, the quicker it incredibly is falling, e.g. whilst the temperature is one hundred deg (i.e. 60 above) it incredibly is falling two times as quickly as whilst it incredibly is 70 deg (30 above), and three times as whilst it incredibly is 60 deg (20 above) combine dT/(T - 40) = -ok dt ln(T - 40) = -kt + const. whilst t = 0, T = one hundred, consequently const = ln 60 subsequently ln(T-40) - ln 60 = -kt ln((T-40)/60) = -kt (T-40)/60 = e^(-kt) whilst t = 3. T = ninety consequently e^(-3k) = 5/6, ok = (-a million/3) ln(5/6) substitute T = 60 into ln((T-40)/60) = -kt: -kt = ln(20/60) = ln(a million/3) consequently t = [ln(a million/3)] / [(a million/3)ln(5/6)] . = 18.077 So it cools to 60 deg in 18 minutes (or approximately 1085 s to the closest entire type of seconds)