Continuity of multiplied 2 functions.

[suhcrates]

I'll introduce you first the 'given question' I found on a math book.
And I'm curious if this question even makes sense, and this is the main part of my post. I wrote 'my question' below.

Given question:
[[[[[ There's 2 functions.

f(x)=x^2 - 4x + a
g(x)= { 2 if (|x-b|>1)
1 if (|x-b| <=1 }

f(x)*g(x) is 'continuous' in whole set of real number.
And what's the constant number a, b ?
]]]]]]]]]

the book saies the answer is " a=-2, b=1 "


But my question:
Is it even possible that f(x)*g(x) be 'continuous'?
g(x) is not continuous, and if you check out how g(x) looks like, it seems f(x)*g(x) can't be continuous neither.


this is what g(x) may look like.



And this is f(x)*g(x) may look like.



Isn't these graphs correct? Then no matter what a, b is, isn't it impossible for f(x)*g(x) to be continuous?
This function have no 'Limit->b-1 : f(x)g(x)' nor 'Limit -> b+1 : f(x)g(x)'. Because after and before the value of x=b-1 or x=b+1, the y values are segmented.

 

[Dr.Peterson]

Something special must be true in order for the product function to be continuous; clearly it will not be so in general -- that's the point of the problem! So don't be focusing on the general case.

Did you look at what the claimed solution looks like? That should give you some idea of what they are doing.

You will have f(x) = x^2 - 4x - 2, and g(x) will jump at 0 and 2. What will f(x)*g(x) look like?

Unfortunately, I think their answer is wrong, so looking at it is not helpful. Just think about what it would take to make the limits exist. What could you multiply by to make the left- and right-hand limits the same?

 

[JeffM]

It might have been a better problem had it said "Are there any values of a and b where h(x) is continuous given

[MATH]h(x) = f(x) * g(x),\ f(x) = x^2 - 4x + a,\ g(x) = 2 \text { if } |x - b| > 1, \text { and } g(x) = 1 \text { if } |x - b| \le 1.[/MATH]
It is obvious that f(x) is continuous for all real x, and h(x) is continuous at all x except where x = b + 1 and x = b - 1.

[MATH]m(x) = x^2 - 4x + a \implies[/MATH]
[MATH]m(b + 1) = b^2 + 2b + 1 - 4b - 4 + a = b^2 - 2b + a - 3 \text { and}[/MATH]
[MATH]m(b - 1) = b^2 - 2b + 1 - 4b + 4 + a = b^2 - 6b + a + 5.[/MATH]
It should be obvious that m(x) is continuous at all x.

[MATH]n(x) = 2x^2 - 8x + 2a \implies n(b + 1) = 2b^2 - 4b + 2a - 6 \text { and } n(b) = 2b^2 - 12b + 2a + 10.[/MATH]
It should be obvious that n(x) is continuous at all x.

If there exist values of a and such that m(b + 1) = n(b + 1) and m(b - 1) = n(b - 1), what would that entail about h(x)?

 

[HallsofIvy]

g(x) is clearly not continuous at x= 1. Something must happen to f(x) at x= 1 that "wipes out" that discontinuity at x= 1. Recall that 0 times any number is 0.