Thread: concavity...why?

I have a few problems that i don't understand the answer..

f(x)=2x^2 -3x+4
f'(x)= 4x-3
f"(x)=4
answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...

f(x)=x^4/7
f'(x)= 4/7 x^-3/7
f"(x)=-12/49 x^-10/7
answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)

f(x)= root 4-x = (4-x)^1/2
f'(x)=1/2 (4-x)^-1/2
f"(x)=-1/4 (4-x)^-3/2
amswer s cd (-infinity,4)......if i plug in say f(5) i'd get a positive..so why is it not cd(-infinity,4) AND cu (4, infinity)?

thanks a bunch...i really appreciate u guys helping me, i only have bits and pieces of this stuff..but i'm getting the bigger picture a little at a time

Originally Posted by larasanut

I have a few problems that i don't understand the answer..

f(x)=2x^2 -3x+4
f'(x)= 4x-3
f"(x)=4
answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...

Mr F says: Because the domain of f is all real numbers and f'' > 0 for all real numbers.

f(x)=x^4/7
f'(x)= 4/7 x^-3/7
f"(x)=-12/49 x^-10/7
answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)

Mr F says: Because the domain of f is all real numbers and f'' < 0 for all real numbers except x = 0 (where it's undefined).

f(x)= root 4-x = (4-x)^1/2
f'(x)=1/2 (4-x)^-1/2
f"(x)= -1/4 (4-x)^-3/2
amswer s cd (-infinity,4)......if i plug in say f(5) i'd get a positive Mr F says: That's strange. Especially since x = 5 is not in the domain of f since f(5) is not defined. Concavity can obviously only exist where the fuction exists, that is, over the domain of the function.

..so why is it not cd(-infinity,4) AND cu (4, infinity)?

thanks a bunch...i really appreciate u guys helping me, i only have bits and pieces of this stuff..but i'm getting the bigger picture a little at a time

Your problems are a combination of carelessness and a lack of understanding of the defintion of concavity.
Read this to address the latter problem: Concavity and Points of Inflection

Originally Posted by larasanut

I have a few problems that i don't understand the answer..

$\displaystyle f(x)=2x^2 -3x+4$
$\displaystyle f'(x)= 4x-3$
$\displaystyle f"(x)=4$
answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...

You have found the derivatives correctly. So we know that if $\displaystyle f(x)$ on $\displaystyle [a,b]$ this implies that $\displaystyle f''(x)>0\quad\forall{x}\in[a,b]$

So what interval is $\displaystyle f(x)=4>0$? It is obviously $\displaystyle (-\infty,\infty)$. Or in other words every real number.

$\displaystyle f(x)=x^4/7$
$\displaystyle f'(x)= 4/7 x^-3/7$
$\displaystyle f"(x)=-12/49 x^-10/7$
answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)

Consider this $\displaystyle x^{\frac{-10}{7}}=\frac{1}{x^{\frac{10}{7}}}=\left(\frac{1}{ x^{\frac{5}{7}}}\right)^2$

So we know that this will be greater than zero forall values of x. So ifwe multiply it by a negative number it will be less than zero for all x.

$\displaystyle f(x)= \sqrt{4-x} = (4-x)^{\frac{1}{2}}$
$\displaystyle f'(x)=\frac{1}{2} (4-x)^{\frac{-1}{2}}$
$\displaystyle f"(x)=\frac{-1}{4} (4-x)^{\frac{-3}{2}}$
amswer s cd $\displaystyle (-\infty,4)\$......if i plug in say $\displaystyle f(5)$ i'd get a positive..so why is it not cd $\displaystyle (-\infty,4)$ AND cu (4, infinity)?

Your analysis would be correct except the values greater than four are imaginary.

Mathstud..you helped me out tremendously on the first and last...but they second one i'm still confused on...i get its not defined at 0..but i still don't get why its concaved down...

Originally Posted by larasanut

I[snip]
f(x)=x^4/7

f'(x)= 4/7 x^-3/7

f"(x)=-12/49 x^-10/7

answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)
[snip]

Originally Posted by larasanut

Mathstud..you helped me out tremendously on the first and last...but they second one i'm still confused on...i get its not defined at 0..but i still don't get why its concaved down...

Try reading my earlier reply (especially the bit about carelessness)!

Note that $\displaystyle (-1)^{-10/7} = \frac{1}{(-1)^{10/7}} = \frac{1}{[(-1)^{1/7}]^{10}} = (-1)^{10} = 1$. Therefore $\displaystyle f''(-1) = -\frac{12}{49} \cdot (1) < 0$.

You need to be much more careful with basic arithmetic.

i get that if it was 1/(-1)^10/9...but i don't see how that is the same as -12/49(-1)^10/9...neg times a neg is pos...so why is it not upwarad

i got...dur...

i mean i got it so if it was say -12/49(-1)^-17/9 it would be upward

Originally Posted by larasanut

i get that if it was 1/(-1)^10/9...but i don't see how that is the same as -12/49(-1)^10/9...neg times a neg is pos...so why is it not upwarad

*Sigh*

Originally Posted by mr fantastic

[snip]
Note that $\displaystyle (-1)^{-10/7} = \frac{1}{(-1)^{10/7}} = \frac{1}{[(-1)^{1/7}]^{10}} = (-1)^{10} = {\color{red}1}$.
[snip]

Note very carefully. It's 1. NOT -1. When a negative number is raised to an even power you get a positive number. Not a negative number.

So you have neg times pos = neg.

It's all there in my previous reply if only you would take greater care in reading and processing it.