Find the concavity and point of inflection(s).
How do you find the second derivative?
Calculus - Reiny, Friday, February 12, 2010 at 8:04am
First of all, I got y' to be
I use x^(1/2) in my steps for √x, I am sure you do too.
I then changed
y' = 1/((2√x(√x+1)^2) to
y' = ((2√x(√x+1)^2)^-1 and used the chain rule
It got a bit messy, but I hope you can end up with
y'' = -(3√x + 1)/((4x^(3/2)(√x+1)^3)
for points of inflection set this to zero and solve
Of course we can forget about the denominator, so
3√x + 1 = 0
3√x = -1
x = 1/9
But since we squared we have to check our answer.
subbing that back into y'' does not give us a zero,
so there is no point of inflection.
Looking at the original equation, it is obvious that the curve starts at (0,0), since x >= 0
continues to rise slowly, and has a limit of y = 1
There is no maximum or minimum point, nor is there a point of inflection.
Calculus - sh, Friday, February 12, 2010 at 10:25am
So my first derivative is wrong?
Is the first step to the second derivative,
y"=-(2√x(x+1)^2)^-2 (-1) ?
Calculus - Reiny, Friday, February 12, 2010 at 10:38am
Yes, your first derivative is wrong,
the correct one I stated above
change that to
y' = [(2√x(√x+1)^2]^-1
use x^(1/2) for √x
and apply the chain rule.
It is unfortunate that we have to do all that messy math to come to the conclusion which is clearly obvious by looking at the graph.
Calculus - sh, Friday, February 12, 2010 at 10:54am
I did use x^(1/2).
I got y'=[1/2x^(-1/2)]/(√x+1)^2,
then I brought the top to the bottom.
The last part of the question is sketching it. The graph is basically f(x)=√x with a horizontal asymptote at y=1.
Thank you Reiny :)
Calculus - Reiny, Friday, February 12, 2010 at 11:07am
Your line of I got y'=[1/2x^(-1/2)]/(√x+1)^2 was correct, but when you brought it to the bottom,
the (1/2) in front of 1/2x^(-1/2) is not governed by the exponent of (-1/2)
so it merely stays as a 1 in the numerator and a 2 in the denominator.
Your graph is correct.