Find all intervals on which the graph of f(x)=(x-1)/(x+3) is concave upward

To find concavity we need to find the second derivative.
Use quotient rule to find f'(x)=4/(x+3)^2
Then use chain rule to find f''(x)=-8/(x+3)^3
To find potential inflection points, we need to find all x values where the second derivative is equal to 0 or is undefined.

Set the numerator and denominator equal to 0 and solve.

(x+3)^3=0
x+3=0
x= -3

Now plug a value less than -3 and greater than -3 into the second derivative to find where the concavity is upward (positive number).

Since f''(-4) is positive, that means that f(x) is concave up on the interval (-infinity, -3).

The answer of this question is 672 .