Minimum of $\phi(x)=\frac1{8x-1}+4x^2-x$ 
Let $\phi:\mathbb{R}\to\mathbb{R}$ be given by $\phi(x)=\dfrac{1}{8x-1}+4x^2-x\;\forall x\in(\frac{1}{8},\infty)$. Find the minimum of $\phi$.

I haven't been able to solve this problem by finding the zeros of $\phi'(x)$ and as that's the only method I know to find the minimum of a function, I'd be very glad if someone could help me with this.
Thank you very much in advance.
 A: $$\phi(x)=\frac{1}{(8x-1)}+4x^2-x \implies \phi'(x)=\frac{-8}{(1-x)^2}+8x-1 =0 $$ $$\implies
(8x-1)^3=8 \implies 8x=3 \implies x=3/8$$ Next we get
$$\phi''(x)=\frac{192}{(8x-1)^2}+8 \implies \phi''(8/3)>0$$
Hence $\phi(x)$ has a local minimum at $x=3/8$ and $$\phi_{min}=\phi(3/8)=\frac{11}{16}$$
A: $\begin{array}\\
f(x)
&=\dfrac{1}{8x-1}+4x^2-x\\
&=\dfrac{1}{8x-1}+4x^2-x+\dfrac1{16}-\dfrac1{16}\\
&=\dfrac{1-(8x-1)/16}{8x-1}+(2x-\dfrac14)^2\\
&=\dfrac{17/16-x/2}{8x-1}+\dfrac1{16}(8x-1)^2\\
&=\dfrac1{8x-1}-\dfrac1{16}+\dfrac1{16}(8x-1)^2\\
g(y)
&=\dfrac1{y}-\dfrac1{16}+\dfrac1{16}y^2
\qquad y=8x-1\\
g'(y)
&=-\dfrac1{y^2}+\dfrac{y}{8}\\
&= 0
\qquad\text{when } y = 2\\
g(2)
&=\dfrac12-\dfrac1{16}+\dfrac14\\
&=\dfrac{11}{16}
\qquad\text{at } x=\dfrac{y+1}{8} = \dfrac38\\
\end{array}
$
Since $g(y) \to \infty$
for $y \to 0$
and $y \to \infty$,
this is a minimum.
There may be
an algebraic way to show that
$g(y) \ge g(2)$
but I''ll leave it at this.
(Added later)
Aha!
Here's how to do it.
$\begin{array}\\
g(y)-g(2)
&=\dfrac1{y}-\dfrac1{16}+\dfrac1{16}y^2-\dfrac{11}{16}\\
&=\dfrac1{y}-\dfrac34+\dfrac1{16}y^2\\
&=\dfrac{y^3-12y+16}{16y}\\
&=\dfrac{(y-2)^2(y+4)}{16y}\\
&\ge 0
\qquad\text{for } y \ge 0\\
\end{array}
$
