Minimum of $\phi(x)=\frac1{8x-1}+4x^2-x$

Question

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.

Answer 1

$$\phi(x)=\frac{1}{(8x-1)}+4x^2-x \implies \phi'(x)=\frac{-8}{(1-x)^2}+8x-1 =0 $$ $$\implies (8x-1)^3=8 \implies 8x=3 \implies x=3/8$$ 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 $$\phi_{min}=\phi(3/8)=\frac{11}{16}$$

Answer 2

$\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} $