$f(x) = x^2 - 6x + 4$. 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 ?

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score 2 · Answer 1 · answered May 31 '20 at 08:45

2

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

Interpret.

answered May 31 '20 at 08:45

David G. Stork

29,774

score 1 · Answer 2 · answered May 31 '20 at 09:01

1

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

answered May 31 '20 at 09:01

razivo

2,205

score 0 · Answer 3 · answered May 31 '20 at 08:57

0

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

answered May 31 '20 at 08:57

Ravindra

414

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

3 Answers3