find the constant C that makes g continuous on (- infinity, infinity)?

Question

g (x) = x^2 - c^2, if x < 4
cx + 20, if x >= 4

Answer 1

In order for g to be continuous, at the point where it switches from one equation to the other, the value of each equation must be the same.

So you want to solve:

x^2 - c^2 = cx + 20

... when x=4:

4^2 - c^2 = 4c + 20
16 - c^2 = 4c + 20
c^2 + 4c + 4 = 0

(c+2)^2 = 0
c = -2

Answer 2

The only point g(x) could be possibly discontinuous is x = 4.

If g(x) *is* continuous there, then:

x² - c² = cx + 20.

So, setting x = 4, gives:

16 - c² = 4c + 20;

or, after rearranging,

c² + 4c + 4 = 0.

After a little thought, this easily factors into:

(c + 2)(c + 2) = 0,

the only real solution to this being, c = ?

Makes sense?
~Soylent Yellow

Answer 3

Since this is a piecewise fuction, the two peices need to match up when x = 4 to produce a continous function.
To do that you must set the two function equal to each other when x = 4 because that is the specified point:
x^2 - c^2 = cx + 20
4^2 - c^2 = 4c + 20
16 - c^2 = 4c + 20
0 = c^2 + 4c + 4 Now factor
0 = (c + 2)^2
c = -2

Answer 4

At the point x = 4, you want x^2 - c^2 = cx + 20
Substituting x =4, you want 16 - c^2 = 4c + 20

So, c^2 + 4c + 4 = 0 or (c + 2)^2 = 0.
Thus c = -2