Piecewise Function: Finding Constants for Continuity

[lomantak]

Hello,

Here is a piecewise function that I came over, and it does not seem to have a definite answer, and so I beg of your recondite knowledge to guide me on this one:f(x) =
ax^2 + bx + c if -oo < x < 0
d if x = 0
[(x^2)(sin(1/x))]-2 if 0 < x < oo

Find all values of the constants a, b, c and d that make the function f continuous on -oo < x < oo.

I think a, b and c are all reals, and d is -2, but I am not sure...

 

[mathman]

For continuity you need c=d=-2. a and b could be anything.

 

[tacman]

Hello,

Thank you for sharing this piecewise function with us. Finding the constants for continuity in a piecewise function can be a bit challenging, but with the right approach, we can find the correct values for a, b, c, and d.

First, let's define what it means for a function to be continuous. A function is continuous if it has no gaps or breaks in its graph. In other words, the function's value at a certain point should be equal to the limit of the function as it approaches that point from both sides.

Now, let's look at the function f(x) that you have provided. We need to find the values of a, b, c, and d that will make this function continuous on the interval -oo < x < oo. To do this, we need to consider the following cases:

Case 1: x < 0
In this case, the function is equal to ax^2 + bx + c. For the function to be continuous, the value of this expression at x = 0 should be equal to the limit of the function as x approaches 0 from the left. This means that we need to have:

ax^2 + bx + c = lim (x->0-) f(x)
ax^2 + bx + c = lim (x->0-) [(x^2)(sin(1/x))]-2
ax^2 + bx + c = 0 - 2 = -2

Case 2: x > 0
In this case, the function is equal to [(x^2)(sin(1/x))]-2. Similarly, for the function to be continuous, the value of this expression at x = 0 should be equal to the limit of the function as x approaches 0 from the right. This means that we need to have:

[(x^2)(sin(1/x))]-2 = lim (x->0+) f(x)
[(x^2)(sin(1/x))]-2 = lim (x->0+) [(x^2)(sin(1/x))]-2
[(x^2)(sin(1/x))]-2 = -2

Since we have the same value for both cases, we can conclude that d = -2. Now, let's consider the first case again and substitute d = -2:

ax^2 + bx + c = lim (x->0-) f(x)
ax