I have $\displaystyle f(x)= x^4 + 2x^2$
and am trying to solve
$\displaystyle f'(x)= 1$, that is $\displaystyle 4x^3 + 4x= 1$
I have $\displaystyle f(x)= x^4 + 2x^2$
and am trying to solve
$\displaystyle f'(x)= 1$, that is $\displaystyle 4x^3 + 4x= 1$
Ok, no, that's not possible.
Bring all the terms on the same side.
$\displaystyle 4x^3 + 4x= 1$
$\displaystyle 4x^3 + 4x- 1=0$
However, this doesn't seem to have three real roots.
So, we'll go by iteration.
Set one x as subject of formula.
$\displaystyle 4x = 1-4x^3$
$\displaystyle x = \dfrac{1-4x^3}{4}$
$\displaystyle x_{n+1} = \dfrac{1-4x_n\ ^3}{4}$
Lets take... x = 0.5
$\displaystyle x_1 = 0.125$
$\displaystyle x_2 =0.248$
$\displaystyle x_3 = 0.245$
.
.
.
$\displaystyle x \approx 0.236$