Respuesta :
[tex]f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
[/tex]
It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
[tex]\lim_{x \rightarrow 4} \ f(x) = f(4)[/tex]
In notation we write respectively
[tex]\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)[/tex]
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
[tex]\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2[/tex]
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²
[tex]c^2+4c+4=0 \\(c+2)^2=0 \\c=-2[/tex]
It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
[tex]\lim_{x \rightarrow 4} \ f(x) = f(4)[/tex]
In notation we write respectively
[tex]\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)[/tex]
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
[tex]\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2[/tex]
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²
[tex]c^2+4c+4=0 \\(c+2)^2=0 \\c=-2[/tex]
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers [tex](-\infty, +\infty)[/tex]
