Since the denominator in f'(x) is (2x-4)^2, for any value of x> 2, f'(x) is negative. So the function is decreasing for any value of x> 2.
the rational function is (x+5)/(2x-4)
I have solved for the first and second derivatives and they are
f'(x)= -14/(2x-4)^2
f''(x)= 56/(2x-4)^3
In my text it gives an example of a similar question and its points of increase and decrease but it says nothing about how you solve for it using a rational function. If anyone can explain the process to me I would appreciate it. Thank you
Presumably, then, you know that a function is increasing where its derivative is positive and decreasing where its derivative is negative.
It should be easy to see that f'(x)= -14/(2x-4)^2 is NEVER positive because (2x-4)^2, being a square, is never negative.
Similarly a function is "concave upward" when its second derivative is positive and concave downward when its second derivative is negative. Since now 2x- 4 is cubed, f" will be positive when 2x- 4> 0 and negative when 2x- 4< 0.
Finally, a "point of inflection" occurs where the second derivative is 0- since a fraction is only 0 when its numerator is 0, that doesn't happen here. This function changes concavity on where it is not defined.