Thread: Identify the discontinuity. f'(x) for the f(x).

1. Identify the discontinuity. f'(x) for the f(x).

I did graph the inequality.

Is see the y intercept at 5 and what looks like opposite slopes. I replaced a with 5. The notation of question e):

$\displaystyle f'(x)$ for the $\displaystyle f(x)$ in d)

is a little bit confusing to me. If I am correct the $\displaystyle f'(x)=-1$ and $\displaystyle f'(x)=1$ for the function in d)

I am trying to determine how to graph:
$\displaystyle f'(x)$ for the $\displaystyle f(x)$ in d)

2. Re: Identify the discontinuity. f'(x) for the f(x).

Part d is a v shape Just the top part of your graph. Ofcourse, you don't know what the value of 'a' is so you don't know if the vertex is above or below the x axis, but you do know it is a V
When x>0 the gradient is +1, when x<0 the gradient is -1
What happens when x=0? The graph is not defined. There is no f(x) value when x=0, f(x) is undefined when x=0 so f(x) is discontinuous when x=0
You show this on a graph by putting an open circle around this point (0,a)

Now when x>0 f'(x)=1 This is just a line horizontal to the x axis starting at (0,1) x=0 is not included so it will start with an open circle
And, when x<0 f'(x)=-1 This is just a line horizontal to the x axis starting at (0,-1) x=0 is not included so it will start with an open circle

So f'(x) is discontinuous at x=0 and the two halves start in very different places.

(Sorry, I don't know what the words are for different types of discontinuity)