Fuctions, Interval, Increasing/ Decreasing, Concave up/ Concave Down (Using Calculus)

Question

Consider the function f(x)=3x+6/6x+3

                                              

For this function there are two important intervals: (−∞,A) and (A,∞) where the function is not defined at A.

 

Find A: ___

 

For each of the following intervals, tell whether f(x) is increasing or decreasing.

(−∞,A): ___
(A,∞): ___

 

This function has no inflection points, but we can still consider its concavity. For each of the following intervals, tell whether f(x) is concave up or concave down.

(−∞,A): ___
(A,∞): ___

Answer 1

I am assuming you mean (3x+6)/(6x+3).

 

A rational function(a fraction) is defined everywhere except when the denominator is 0....that is where it has a vertical asymptote.

 

So denominator = 0 when 6x+3=0 which is when x=-1/2.

So this is continuous from(-infinty,-1/2) and (1/2,infinity).

 

The first derivative requires the quotient rule

f'(x) = [ (6x+3)(3) - (3x+6)(6) ] / (6x+3)² = -27/(6x+3)²

The numerator is always negative and the denominator, since it's squared, is always positive except at -1/2. So the first derivative is always negative, which means f(x), the original function is always decreasing.

 

c)concavity is determined by the second derivative. It is easier to rewrite the f'(x) = -27(6x+3)^-2 and take the derivative using the chain rule.

 

f''(x) = 54(6x+3)^-3(6) = 324/(6x+3)^3.

Plug in something in each interval and determine if it's positive or negative.

f''(-1) = 324/(-27). Since this is negative, it is always concave down on this interval.

f''(0) = 324/27. Since this is positive, it is concave up on the second interval.

 

Hope this helps.