# Finding max mins point of inflection.

• Sep 4th 2009, 05:30 PM
el123
Finding max mins point of inflection.
$f(x)= 2x^2 + \frac {8}{x^2}$

$f'(x)= 4x - \frac{16}{x^3}$

$f''(x) = 4+\frac{48}{x^4}$

So taking f''(x)

$\frac{4}{1} + \frac{48}{x^4}=\frac{4x^4}{x^4}+\frac{48}{x^4}$

$\frac{4x^4+48}{x^4} \cdot x^4 = 0 \cdot x^4$

$4x^4+48 = 0$

$48=-4x^4 => -12 = x^4 => -1.86 = x$

so i have a point of inflection at -1.86?

then evaluate f(-1.86).

$2\cdot -1.86^2 + \frac{8}{-1.86^2} = -9.23$

So this is a local maximum?

somthing seems wrong , can anyone point out what it is?
• Sep 4th 2009, 05:42 PM
skeeter
Quote:

Originally Posted by el123
$f(x)= 2x^2 + \frac {8}{x^2}$

$f'(x)= 4x - \frac{16}{x^3}$

$f''(x) = 4+\frac{48}{x^4}$

So taking f''(x)

$\frac{4}{1} + \frac{48}{x^4}=\frac{4x^4}{x^4}+\frac{48}{x^4}$

$\frac{4x^4+48}{x^4} \cdot x^4 = 0 \cdot x^4$

$4x^4+48 = 0$

$\textcolor{red}{4x^4 + 48 = 0}$ has no real solutions.

f''(x) > 0 for all x in the domain of f, so f is concave up everywhere in its domain, i.e. no inflection points.

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• Sep 4th 2009, 05:48 PM
el123
is that because $\sqrt[4]{-12}$ doesnt exist?

Are you able to tell me or direct me to somewhere where i may learn the process of finding max/mins , i have been trying to find somwhere but i can't. My textbook is worthless.Im getting really frustrated trying to work these out i just don't understand them.
• Sep 4th 2009, 05:50 PM
el123
is that because $\sqrt[4]{-12}$ doesnt exist?

Are you able to tell me or direct me to somewhere where i may learn the process of finding max/mins , i have been trying to find somwhere but i can't. My textbook is worthless.Im getting really frustrated trying to work these out i just don't understand them.