Question

Help Please!
Find the value of the constant c that makes the following function continuous on (-infinity, +infinity)

f(x)={ x^2-c if -infinity 12 years ago

Answer 1

@tkhunny

Answer 2

u have
\[\large f(x)=\begin{cases}
x^2-c, &x<5\\ cx+5, & x\geq 5
\end{cases} \]

Answer 3

use the definition of continuity
\[\large \lim_{x\to5}f(x)=f(5) \]

Answer 4

and the existence condition for limits: both lateral limits must exist and
\[\large \lim_{x\to5+}f(x)=\lim_{x\to5-}f(x) \]

Answer 5

\[f(x)=\left[\begin{matrix}x^2-c, & if -\infty<x<5 \\ cx+5, & if x \ge5\end{matrix}\right]\]
I understand what you've explained but how would I find c.?

Answer 6

@helder_edwin

Answer 7

@Loser66 help?(:

Answer 8

ok. for the left limit, since \(x\to5-\) then \(x<5\) so
\[\large \lim_{x\to5-}f(x)=\lim_{x\to5-}(x^2-c)=5^2-c=25-c \]
for the right limit, since \(x\to5+\) then \(x>5\) so
\[\large \lim_{x\to5+}f(x)=\lim_{x\to5+}(cx+5)=5c+5. \]

Answer 9

now, for the function to be continous, these limits have to be equal so
\[\large 25-c=5c+5 \]
\[\large 20=6c \]
\[\large c=\frac{20}{6}=\frac{10}{3} \]

Answer 10

oh! that made so much sense, I kinda over thought it. Thank you!!! @helder_edwin

Answer 11

u r welcome