Is f(x)=x/sqrt(x+3) f(x)=xx+3 increasing or decreasing at x=5 ?

1 Answer
Dec 11, 2015

Increasing.

Explanation:

Find the first derivative of the function.

If f'(5)<0, then f(x) is decreasing when x=5.
If f'(5)>0, then f(x) is increasing when x=5.

To find f'(x), use the quotient rule.

f'(x)=(sqrt(x+3)d/dx[x]-xd/dx[sqrt(x+3)])/(sqrt(x+3))^2

Find each derivative separately.

d/dx[x]=1

The following derivative requires the chain rule.

d/dx[sqrt(x+3)]=1/2(x+3)^(-1/2)overbrace(d/dx[x+3])^(=1)=1/(2sqrt(x+3))

Plug the derivatives back in.

f'(x)=(sqrt(x+3)-x/(2sqrt(x+3)))/(x+3)

Multiply everything by 2sqrt(x+3) to clear out the denominator of the internal fraction.

f'(x)=(2(x+3)-x)/(2(x+3)sqrt(x+3))

f'(x)=(x+6)/(2(x+3)^(3/2))

Now, find f'(5).

f'(5)=(5+6)/(2(5+3)^(3/2))=11/(2(8)^(3/2))

We could determine the exact value of f'(5), but it should be clear that the answer will be positive.

Thus, f'(x) is increasing when x=5.