Question

f(x) = ((ln(7x−x^2)/ 12))^3/2 is it an increasing function

Answer 1

f(x ) = { ln(7x - x²)/12}^3/2

f(x ) is increasing when , f'(x ) > 0

now , differentiate wrt x
df(x )/dx = 3/2{ ln(7x - x²)/12 }½ × 1/12(7x -x²) × ( 7 - 2x )

= {3/2 × 1/12} × √ln(7x - x²)/√12 ×( 7 -2x)/(7x - x²)

={ 3/24 × 1/2√3 } × √ln(7x - x²) × ( 7 -2x)/( 7x - x²)


f'( x ) > 0 when, ( 7 -2x)/( 7x - x²) >0 and squre root function always positve

( 7 -2x )/x ( 7 -x) > 0

after solving this we get ,
x€ ( 0, 7/2) U( 7, ∞) ------(1)

now , for log to be defined ,
( 7x - x²) > 0
x( 7 - x) > 0

0 < x < 7 -----(2)

also for √ln( 7x -x²)
ln( 7x - x²) ≥ 0
( 7x - x²) ≥ 1
x² - 7x +1 ≤ 0
{ x - (7 ±3√5)/2}≤ 0

(7 -3√5)/2 ≤ x ≤ (7 + 3√5)/2 -------(3)

take common value of x from (1), (2) and (3)


★ ★ x € [ (7-3√5)/2 , (7+3√5)/2] ★★

hence f(x ) is increasing above interval ,