# `f(x) = (x + 3)/(sqrt(x))` Find the points of inflection and discuss the concavity of the graph of the function.

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### 2 Answers

Given the function `f(x)=(x+3)/sqrt(x)`

i.e it can be written as `f(x)=x/sqrt(x)+3/sqrt(x)` =`x^(1/2)+3x^(-1/2)`

Taking the first derivative we get,

`f'(x)=1/2x^(-1/2)-3/2x^(-3/2)`

Now again differentiating we get,

`f''(x)=-1/4x^(-3/2)+9/4x^(-5/2)`

Now theĀ points of inflection can be found out by equating f''(x) to zero. i.e.

`f''(x)=0`

i.e. `-1/4x^(-3/2)+9/4x^(-5/2)=0`

`9x^(-5/2)-x^(-3/2)=0`

`x^(-3/2)(9/x-1)=0`

`x^(-3/2)((9-x)/x)=0`

`(9-x)/x^(5/2)=0`

Therefore x=9 is the inflection point.

Now choose auxilliary points x=8 to the left of inflection point and x=10 to the right of inflection point.

So`f''(8)=0.0013>0`

Therefore on (-infinity, 9) , the curve is concave upward .

Now,

f''(10)=-0.0007<0

Therefore on (9, infinity) , the curve is concave downward.

Find the second derivative.

Therefore, given

`f(x)=(x+3)/sqrtx `

Then,

`f'(x)=(x-3)/(2x^(3/2)) `

and

`f''(x)=-(x-9)/(4x^(5/2)) `

Now we need to find the value(s) of x that make the second derivative zero:

`0=-x+9 `

`x=9 `

Try x values on both sides, like 8 and 10

`f''(8)=0.00138 `

`f''(10)=-7.91*10^-4 `

The concavity does change so `x=9 ` is a point of inflection.

Find the y-value, `(9,4) `

Therefore,

`(9,4) ` is a point of inflection