Let f(x)=x^2-8x+16 and g(x) x^2-16. For what values of x is f(x)/g(x) increasing?

Question

Explain plz~

Answer 1

Let h(x) = f(x)/g(x)

h is increasing when h' > 0

h' = (f'g-fg')/g^2
= [(2x-8)(x^2-16) - (x^2-8x+16)(2x)]/(x^2-16)^2
= 8/(x+4)^2

h' > 0 for all real x ≠ -4
so, h is increasing for all real x ≠ 4 or -4

The calculation is much easier if you note that
f = (x-4)^2
g = (x-4)(x+4)

f/g = (x-4)/(x+4) = 1 - 8/(x+4)
so,
(f/g)' = 8/(x+4)^2