Find all values of c such that f is continuous on (-∞, ∞). $ f(x) = \left\{ \begin{array}{lcl} {\color{red}5} - x^{2}, & & x \le c \\ x, & & x > c \end{array}\right. $

[tex]f(x)=\begin{cases}5-x^2&\text{for }x\le c\\x&\text{for }x>c\end{cases}[/tex]

will be continuous as long as

[tex]\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)[/tex]

We have

[tex]\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c}(5-x^2)=5-c^2[/tex]

[tex]\displaystyle\lim_{x\to c^+}f(x)=\lim_{x\to c}x=c[/tex]

so we must satisfy

[tex]5-c^2=c\implies c^2+c-5=0\implies c=\dfrac{-1\pm\sqrt{21}}2[/tex]

Answer Link

MrRoyal MrRoyal · Answer 2 · 2022-02-04T14:55:09+01:00

The values of c such that the function f is continuous are: c = -2.79 and c = 1.79

The function is given as:

[tex]f(x) = \left\{ \begin{array}{lcl} {5} - x^{2}, & & x \le c \\ x, & & x > c \end{array}\right.[/tex]

For the function to be continuous, then the following must be true

[tex]5 - x^2 = x[/tex]

Substitute c for x

[tex]5 - c^2 = c[/tex]

Express as a quadratic function

[tex]c^2 + c - 5 = 0[/tex]

Using a graphing calculator, the values of c are:

c = -2.79 and c = 1.79

Read more about continuous functions at:

https://brainly.com/question/9412728