What value of c, if any is the function continuous at x=3?

[onesun0000]

At first, I equated the first and last pieces of the function, plugged in 3 for x and got 6. But then I realized that neither are defined at 3 so it made me think that maybe the answer is that there's no such value for c that will make the function continuous since it'll still be undefined at x=3 for the first and last pieces. Which is correct?

 

[skeeter]

you're working with limits here ...

[MATH]f(3)=7 \implies \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = 7[/MATH]
[MATH]c+3c-9 = 7[/MATH]
[MATH]2c+ \dfrac{3}{3-2} = 7[/MATH]
solve the system

 

[onesun0000]

you're working with limits here ...

[MATH]f(3)=7 \implies \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = 7[/MATH]
[MATH]c+3c-9 = 7[/MATH]
[MATH]2c+ \dfrac{3}{3-2} = 7[/MATH]
solve the system

The system is inconsistent. So there must not be a value for c that makes the function continuous.

 

[skeeter]

IMHO, no solution on a multiple choice question shows laziness in writing the question. The author should have tried using two constants to solve for instead.

 

[HallsofIvy]

You would need to have, as skeeter said in the first response to you question, $\displaystyle c+ c(3)- (3)^3= 4c+ 9= 7$ so 4c= -2, c= -1/2 and $\displaystyle 2c+ \frac{3}{3- 2}= 2c+ 3= 7$ so 2c= 4, c= 2.

c can't have both those values so there is no such value of c. That is answer (D). I don't know what you mean by "undefined for x= 3".