Math Help - Concavity problem...what am i doing wrong?

1. Concavity problem...what am i doing wrong?

I get the concavity backwards....I get cu (- infinity,1) cd (1, infinity)
but the answer says cd (-infinity, 1) cu (1, infinity)...I might just be getting the derivitives wrong

The problem is t2/t-1

For f'(x) I get -t^2+2t/(t-1)^2 or -t(t-2)/(t-1)^2
For f''(x) I get -2(2t+1)/(t-1)^3

I set f''(x) equal to zero...i use the denominator right? (t-1)= 0...t=1

For the critical number i get 1...plug in my test points and i get it wrong..so i figure it has to be my derivatives...Can someone work out the whole problem so i can see where i'm messing up..thanks

2. Originally Posted by larasanut
I get the concavity backwards....I get cu (- infinity,1) cd (1, infinity)
but the answer says cd (-infinity, 1) cu (1, infinity)...I might just be getting the derivitives wrong

The problem is t2/t-1

For f'(x) I get -t^2+2t/(t-1)^2 or -t(t-2)/(t-1)^2
For f''(x) I get -2(2t+1)/(t-1)^3

I set f''(x) equal to zero...i use the denominator right? (t-1)= 0...t=1

For the critical number i get 1...plug in my test points and i get it wrong..so i figure it has to be my derivatives...Can someone work out the whole problem so i can see where i'm messing up..thanks
I'm not sure that I understand the question correctly. If you mean:

$f(t)=\dfrac{t^2}{t-1}$

Then the first derivation is:

$f'(t)=\dfrac{(t-1)\cdot 2t-t^2}{(t-1)^2}=\dfrac{t(t-2)}{(t-1)^2}$

3. Originally Posted by larasanut
I get the concavity backwards....I get cu (- infinity,1) cd (1, infinity)
but the answer says cd (-infinity, 1) cu (1, infinity)...I might just be getting the derivitives wrong

The problem is t2/t-1

For f'(x) I get -t^2+2t/(t-1)^2 or -t(t-2)/(t-1)^2
For f''(x) I get -2(2t+1)/(t-1)^3

I set f''(x) equal to zero...i use the denominator right? (t-1)= 0...t=1 .... No, at t = 1 the function is not defined, because the division by zero is forbidden!

...
....

4. Originally Posted by larasanut
...
For f''(x) I get -2(2t+1)/(t-1)^3

I set f''(x) equal to zero...i use the denominator right? (t-1)= 0...t=1

For the critical number i get 1...plug in my test points and i get it wrong..so i figure it has to be my derivatives...Can someone work out the whole problem so i can see where i'm messing up..thanks
1. The second derivation of f is:

$f''(t)=\dfrac2{(t-1)^3}$

This function has no real zero.

Use the definitions of concavity:

A function is concave down if the second derivation is negative.
A function is concave up if the second derivation is positive.

Examine

$\dfrac2{(t-1)^3}<0~\implies~(t-1)^3<0~\implies~t<1$ ....... f is cd

$\dfrac2{(t-1)^3}>0~\implies~(t-1)^3>0~\implies~t>1$ ....... f is cu

5. can u work out the second deriviative? that is where i'm geting it wrong

6. Originally Posted by larasanut
can u work out the second deriviative? that is where i'm geting it wrong
I'll do it step by step so you can compare your notes and mybe spot your error:

$
f'(t)=\dfrac{t^2-2t}{(t-1)^2}~\implies~f''(t)=\dfrac{(t-1)^2(2t-2)-(t^2-2t)(2(t-1))}{(t-1)^4}=$
$\dfrac{(t-1)^22(t-1)-(t^2-2t)(2(t-1))}{(t-1)^4}$ Cancel:

$f''(t)=\dfrac{2(t-1)^2-2(t^2-2t)}{(t-1)^3} =\dfrac{2t^2-4t+2-2t^2+4t}{(t-1)^3} =\dfrac2{(t-1)^3}$

7. i know i set f"(x) to =0 but what does that mean, am i really setting the numerator or denominator =0

8. Originally Posted by larasanut
i know i set f"(x) to =0 but what does that mean, am i really setting the numerator or denominator =0
To determine the concavity of a function you normally calculate f''(x) = 0 to get the point where the concavity changes (point of inflection).

In your case this method isn't possible because the function has no point of inflection. Therefore you have to use the definition of concavity as I have shown in my previous post.

In general: A quotient is zero if the numerator is zero and the denominator is unequal to zero. The second derivation of your function is a constant (2) which can't be zero.