$f(x) = x^2 - 6x + 4$. What is the maximum value of $\frac{1}{f(x)}$?

Question

The solution to this in my book says $\frac{1}{5}$. But greater values can be achieved by using $f(x)$ values approaching zero, right ?

Answer 1

Here's a graph of $1/f(x)$:

Interpret.

Answer 2

Make use of $|f(x)|\leq M$ then $|\frac{1}{f(x)}|\geq \frac{1}{M}$

Answer 3

Your book confused a maximal value and a maximum, what he intended you to find is where(and at what value) $$(\frac{1}{f(x)})’=0$$ $$-\frac{2x-6}{x^2-6x+4}=0$$ $$x=3$$ $$\frac{1}{3^2-6*3+4}=-\frac{1}{5}$$