Points where f is discontinuous question? -- please help?

Question

Hi. I would like to find out the answer to this question.

Considering the following function defined on [2, 8]

f(x) =

x^2-6*x + 12 if 2 ≤ x ≤ 3 --- these are less than or equal to symbols.

3 / (x-4)^2 if 3 < x < 6

x^2 - 12*x+ 147 / 4 if 6 < x < 8 (to clarify that is 147 divided by 4)

x^2 - 12*x +23 / 4 if x = 8 (to clarify that is 23 divided by 4)

Then what are the the set of all points x in [2, 8] where f is discontinuous?

Thanks for any help received. It is much appreciated.

Answer 1

There is a point at x = 4 where it tends to infinity .

The functions everywhere else are piecewise continuous ie within the ranges [2,3], (3,4), (4,6), (6,8). All you need to do then is look at the edges of each range.

At 3 it is continuous

At x = 6 is continuous since tending to it from below you reach 3/4 and from to x=6 from above you get 3/4.

So the only discontinuity is at x= 8 where from below you reach 37.75 and at x= 8 the function has the value 68.75 which is not the same.

Hope that helps.

Answer 2

Since f(x) is a piecewise function you have to look at each piece, and where each piece connects.

for 2 ≤ x ≤ 3 the piece is a regular polynomial, so there are no potential discontinuities

the check for a discontinuity at x = 3 compare the value at x=3 of the first piece and the second piece to see if they are equal

(1) x^2 - 6x + 12 =&gt; (3)^2 - 6(3) + 12 = 3

(2) 3 / (x - 4)^2 =&gt; 3 / (3 - 4)^2 = 3

They are equal so no discontinuity there

Now check the second piece for a discontinuity by setting the denominator equal to zero

(x - 4)^2 = 0 =&gt; x = 4

since x = 4 is within the domain of this piece, it is a discontinuity of f(x)

Since neither the second piece or the third piece have the less than or equal to sign in the domain, there is no actual value at x = 6 so that is also a discontinuity

the last part to check is to see if the value at x=8 is the same as x = 8 for the third piece

(3) x^2 - 12x + 147/4 =&gt; (8)^2 - 12(8) + 147/4 = 19/4

(4) x^2 - 12x + 23/4 =&gt; (8)^2 - 12(8) + 23/4 = -105/4

so x=8 is also a discontinuity

So the x values where f(x) is discontinuous are x = 4 , 6 , 8

The question asks for the set of points though, so find the y values to go with each x, which you can&#039;t do for x = 6 unless there was a typo, so for that point I would just write x = 6 isn&#039;t in the domain.

Answer 3

f(x) =

= x² - 6x + 12 if 2 ≤ x ≤ 3

= 3 / (x - 4)² if 3 &lt; x &lt; 6

= x² - 12x+ 147 / 4 if 6 &lt; x &lt; 8

= x² - 12x +23 / 4 if x = 8

f will be discontinuous anywhere there is a jump, or if it is undefined

The function starts at x = 2

x² - 6x + 12 is a continuous function so the function is continuous on the interval (2, 3)

when x = 3

f(3) = 3² - 6(3) + 12

= 9 - 18 + 12

= 3

if you look at what happen as you get near 3 from the positive direction:

lim (x → 3) of 3 / (x - 4)²

= 3 / (-1)²

= 3

so it is continuous at x = 3

3 / (x - 4)² is continuous for all x except when (x - 4)² = 0

so there is a discontinuity at x = 4

Next point to check is x = 6 (where the definition of the function changes

f(6) = 3 / (6 - 4)²

= 3 / 4

lim x → 6 of x² - 12x+ 147 / 4

= 6² - 12(6) + 147 / 4

= 3/4 so it is continuous at x = 6

x² - 12*x+ 147 / 4 is continuous for any x so it is continuous between x = 6 and x = 8

f(8) = 8² - 12(8) + 23 / 4

= -26.25

lim (x → 8) x² - 12x+ 147 / 4

= 8² - 12(8) + 147 / 4

= -20.25

So the function has a discontinuity at x = 8

The function ends when x = 8

Answer 4

The functions on number 1, 2, and 4 are always continuous, since any x we will have a f(x)

The function on number 2 is discontinuous when x = 4, since when x = 4, then  f(4) would be a very large number or positive infinity.

Good Luck!