Math Question

9a) f(x) = [e-3x √(2x - 5)]/(6 - 5x)4

ln f(x) = - 3x + (1/2) ln (2x - 5) - 4 ln (6 - 5x)

Taking differentiation on both sides:

f'(x)/f(x) = -3 + 1/(2x - 5) + 20/(6 - 5x)

= [-3(2x - 5)(6- 5x) + (6 - 5x) + 20(2x - 5)]/[(2x - 5)(6 - 5x)]

= [-3(-10x2 + 37x - 30) + 6 - 5x + 40x - 100]/[(2x - 5)(6 - 5x)]

= (30x2 - 76x - 4)/[(2x - 5)(6 - 5x)]

f'(x) = {[e-3x √(2x - 5)]/(6 - 5x)4} {(30x2 - 76x - 4)/[(2x - 5)(6 - 5x)]}

= [e-3x (30x2 - 76x - 4)]/[√(2x - 5) (6 - 5x)4]

b) f(x) = 5x2

ln f(x) = ln 5 + 2 ln x

Taking diff. on both sides:

f'(x)/f(x) = 2/x

f'(x) = 10x

10ai) lim (x → +∞) (x5 - x4)/(x4 - 6x3 + 1)

= lim (x → +∞) (x - 1)/(1 - 6/x + 1/x4)

= +∞

ii) lim (x → +∞) [√(x2 - 3x) - x]

= lim (x → +∞) {[√(x2 - 3x) - x][√(x2 - 3x) + x]}/[√(x2 - 3x) + x]

= lim (x → +∞) [(x2 - 3x) - x2]/[√(x2 - 3x) + x]

= lim (x → +∞) -3x/[√(x2 - 3x) + x]

= lim (x → +∞) -3/[√(1 - 3/x) + 1]

= -3/2

bi) It does not exist since the function tends to different values when x tends to 2 from left and right hand sides.

ii) It does not exist since the lim (x → 4+) f'(x) =/= lim (x → 4-) f'(x)