Increasing And Decreasing Functions - Free MCQ Practice Test with solutions,

f’(x) = 4sin³x cosx - 4cos³x sinx
= - sin4x
As x is in the interval [0, ∏/2] hence 4x is in the interval [0, 2∏]

 f(x) = a^x
Taking log bth the sides, log f(x) = xloga
f’(x)/a^x = loga
f’(x) = a^x loga   {a^x > 0 for all x implies R, 
for loga e<a<1 that implies loga < 0}
Therefore, f’(x) < 0, for all x implies R
f(x) is a decreasing function.

 f'(x) = [(logx).2−2x.1/x](logx)²
= 2(logx−1)/(logx)^2
∴f'(x)>0
⇔logx−1>0
⇔logx>1
⇔logx>loge
⇔x>e
∴f(x) is increasing in (e,∞)

f(x) = -3x + 12
f(0) = -3(0) + 12 = 0 - 12 = 0
f(1) = 9
f(2)= 6
f(3) = 3
f(4) = 0
f(5) = -3

  f(x) = x³ – 3x² + 4x

f’(x) = 3x² – 6x + 4.

f’(x) = 3(x² – 2x + 1) + 1.

=> 3(x-1)2 + 1>0, in every interval of R. Therefore the function f is increasing on R.