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

2 Answers

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nees101 | High School Teacher | (Level 2) Adjunct Educator

Posted on

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,


Now again differentiating we get,


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


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


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



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

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loves2learn | (Level 3) Salutatorian

Posted on

Find the second derivative.

Therefore, given

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


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


`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) `


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