# What are the points of inflection of f(x)=x^7/(4x-2) ?

Jul 16, 2016

$x = \frac{35 \pm \sqrt{217}}{48}$.
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#### Explanation:

$\left(4 x - 2\right) f = {x}^{7}$. Differentiate twice..

$f ' \left(4 x - 2\right) + 4 f = 7 {x}^{6}$

$f ' ' \left(4 x - 2\right) + 8 f ' = 42 {x}^{5}$

Eliminating f' and f,
$f ' ' = \frac{42 {x}^{5} - 8 \left(7 {x}^{6} - \frac{4 {x}^{7}}{4 x - 2}\right)}{4 x - 2}$

f''=0, for x satisfying $42 {x}^{5} - 8 \left(7 {x}^{6} - \frac{4 {x}^{7}}{4 x - 2}\right) = 0$. This leads to

$x = 0 \mathmr{and} \left(42 - 56 x\right) \left(4 x - 2\right) + 32 {x}^{2} = 0$-

The second reduces to $48 {x}^{2} - 70 x + 21 = 9$ giving $x = \frac{35 \pm \sqrt{217}}{48}$.

Now, use the first three equations..

At x=0, all derivatives vanish.

At x=(35+-sqrt 217)/48#, f''' is not 0

So, the points of inflexion are .$x = \frac{35 \pm \sqrt{217}}{48}$.

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