Determine a piecewise function that coincides with f and is continuous at x = 3

[Mel Gibson]

Hello! I'm in Math 141 Precalculus looking for help with this problem.

Work so far:

can be factored into (x-3)(x+7)/(x-3)

cancel out (x-3) to get x+7

--------------------------
I can see that x+7 coincides with the graph of the first equation and has a value at x=3, but I'm confused as to how you format the answer if x+7 is the solution.

I could write that like:

x + 7 if x < 3
x + 7 if x = 3
x + 7 if x > 3

but that seems very redundant which leads me to believe I'm doing something wrong

Thank you in advance for any help!
-Mel Gibson

 

[Dr.Peterson]

I'm not sure why the question is phrased that way (possibly because other problems do require a piecewise function).

I would just give the answer as g(x) = x + 7, which is defined in one piece! It is common in math to use words in "degenerate" ways like this, where "piecewise" means "piecewise, if necessary", and it is not necessary.

Others may have different opinions.

 

[Steven G]

Redefine f(x) = (x^2 + 4x -21)/(x-3) if x is not 3 and 10 if x=3.
My answer is a piecewise function. If you prefer you could replace f(x) above with g(x).
I too agree that it is best just to say g(x) = x+7