How do I find the derivative of this calculus problem dealing with relative extrema?

Question

I am in relative extrema in my calculus class. I have a problem that states find the relative minimum and relative maximum of the following: G(x) = (x-3)^2

                                x-4

I am having problems finding the derivative that the book gives as (x-3)(x-5)

                                                                                                    (x-4)^2

Can you please explain how I get that derivative? Thank you.

Answer 1

Is it G(x) = (x-3)²/(x-4) ?

 

If so, then by the quotient rule, G'(x) = [2(x-3)(x-4) - (x-3)²(1)] / (x-4)²

 

                                                      = [2(x²-7x+12) - (x²-6x+9)] / (x-4)²

 

                                                      = (x²-8x+15)/(x-4)²

 

                                                      = (x-3)(x-5)/(x-4)²

 

G'(x) = 0 when x = 3 or 5.  Note that G(x) and G'(x) are both undefined at x = 4 (the graph of y = G(x) has a vertical asymptote when x = 4).  

 

When x < 3, G'(x) > 0.  So, G(x) is increasing.

When 3<x<4, G'(x) < 0.  So, G(x) is decreasing.

 

When 4<x<5, G'(x) < 0.  So, G(x) is decreasing.

When x>5, G'(x) > 0.  So, G(x) is increasing

 

Since the derivative changes sign from positive to negative at x=3, there is a relative maximum when x = 3.  

 

Since the derivative changes sign from negative to positive at x = 5, there is a relative minimum when x = 5.

 

 

Answer 2

Use the quotient rule:

 

(f/g)' = (f'g - fg') / g²

 

d/dx [(x - 3)² / (x - 4)] = [2(x - 3)(1)(x - 4) - (x - 3)²(1)] / (x - 4)²

 

= [(x - 3) / (x - 4)²] [2(x - 4) - (x - 3)]

 

= [(x - 3) / (x - 4)²] (x - 5) = (x - 3)(x - 5) / (x - 4)²