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Let #f(x)=(5/2)sqrt(x)#. The rate of change of f at #x=c# is twice its rate of change at #x=3#. What is the value of c?

Answer 1

Emilia Aguilar

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!

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Answer 2

Adam Adkins

The value of ( c ) is ( 36 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.