We start by differentiating, using the product rule and the chain rule.
Let #y = u^(1/2)# and #u = x#.
#y' = 1/(2u^(1/2))# and #u' =1#
#y' = 1/(2(x)^(1/2))#
Now, by the product rule;
#f'(x) = 0 xx sqrt(x) + 1/(2(x)^(1/2)) xx 5/2#
#f'(x) = 5/(4sqrt(x))#
The rate of change at any given point on the function is given by evaluating #x = a# into the derivative. The question says that the rate of change at #x = 3# is twice the rate of change at #x = c#. Our first order of business is to find the rate of change at #x = 3#.
#r.c = 5/(4sqrt(3))#
The rate of change at #x = c# is then #10/(4sqrt(3)) = 5/(2sqrt(3))#.
#5/(2sqrt(3)) = 5/(4sqrt(x))#
#20sqrt(x) = 10sqrt(3)#
#20sqrt(x) - 10sqrt(3) = 0#
#10(2sqrt(x) - sqrt(3)) = 0#
#2sqrt(x) - sqrt(3) = 0#
#2sqrt(x)= sqrt(3)#
#4x = 3#
#x = 3/4#
So, the value of #c# is #3/4#.
Hopefully this helps!
