concavity...why?

**larasanut** · Nov 7th 2008, 01:23 PM

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

**mr fantastic** · Nov 7th 2008, 01:32 PM

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

**Mathstud28** · Nov 7th 2008, 01:34 PM

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...

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

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

$f(x)=x^4/7$
$f'(x)= 4/7 x^-3/7$
$f"(x)=-12/49 x^-10/7$
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 (-)

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

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.

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

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

**larasanut** · Nov 7th 2008, 02:34 PM

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...

**mr fantastic** · Nov 7th 2008, 02:46 PM

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

You need to be much more careful with basic arithmetic.

**larasanut** · Nov 7th 2008, 02:55 PM

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

**larasanut** · Nov 7th 2008, 02:57 PM

i got...dur...

**larasanut** · Nov 7th 2008, 03:00 PM

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

**mr fantastic** · Nov 7th 2008, 03:01 PM

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

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