Question

1.For f(x)=sin^2(x) and g(x)=0.5x^2 on the interval
[-pi/2,pi/2], the instantaneous rate of change of f is greater than the instanteneous rate of change of g for which value of x?
a)0
b)1.2
c)0.9
d)-0.8
e)1.5

2. If g(x)=√(x) (x-1)^(2/3), then the domain of g'(x) is?
a.{x|0<x<1}
b.{x|all real numbers}
c.{x|x ≠0 and x ≠1}
d.{x|0<x}
e.{x|0<x<1or x>1}

Answer 1

1. take the derivative of each, not that bad
If you have a calculator, just plug in each given value and see what happens
or
make a sketch of y = sin^2 (x) and y = (1/2)x^2 from appr -1.6 to +1.6
take a few points using your calculator .

2. Did you get the actual derivative?
I got (7x-3)/(6√x(x-1)^(1/3) )

clearly for the √x , x>0
and for the (x-1)(1/3) we well get anwers for all values of x
But, if x = 1, the (x-1)^(1/3) will give us 0, and we can't divide by zero, so

all positive values of x, x ≠ 1
I suppose e) would describe that