peice wise and continuity

[Nerd187]

I have been working on this problem that I dont entirely understand. The peice wise function is

f(x) = 6.1 square root of ( 1 - (x +.6 / 10 ) ^2 ) -10.6 < x < 0

and the second function is 6.1 square root of ( 1 - (x+.6 / 11.6) ^1.9 , 0 < x< 11

The question here is explain why the function f(x) is continuious at x = 0 and work must be shown. I basically tried and attempted to solve for X but have retained a complicated looking answer.

The second question is the derivative function of f(x) continuous at x = 0 and work must be shown. Here I did the product rule and simplified it as much as I could but im not quite too sure about this one either. Can someone also kinda explain the concept of continuity in the context of this problem? thanks for the help. :wink:

 

[royhaas]

Continuity means f(0) is the same as the limit of f as x approaches zero. But your function isn't even defined at x=0.

 

[-GC-Phoenix]

for this problem you dont need the function to be defined at zero, it is simply a limit problem.
Take the limit of both functions a the right and left side limits.
Therefore if they both have the same limit as they approach zero you can say it is continuous.

 

[soroban]

Hello, Nerd187!

Is there a typo? . . . The function is not continuous at \(\displaystyle x\,=\,0.\)

\(\displaystyle f(x) \;= \;\begin{Bmatrix}6.1\sqrt{1\,-\,(\frac{x\,+\,0.6}{10}})^2 & \;\; & -10.6\,<\,x\,<\,0 \\ \\ \\
6.1\sqrt{1\,-\,(\frac{x\,+\,0.6}{11.6})^{1.9}} & \;\; & 0\,<\,x\,<\,11\end{Bmatrix}\)

The two limits are not equal . . .

\(\displaystyle \lim_{x\to0^-}\,6.1\sqrt{1\,-\,(\frac{x\,+\,0.6}{10})^2} \;=\;6.0890101\)

\(\displaystyle \lim_{x\to0^+}\,6.1\sqrt{1\,-\,(\frac{x\,+\,0.6}{11.6})^{1.9}} \;=\;6.089017324\)