relative extremums

[navigator]

question:- Find relative extremum of f(x)= x+(a-3)/x ;a€R and
Find the points where the function is increasing.

My try: f´(x)= 1-(a-3)/x^2
To find points f´(x)=0
1-(a-3)/x^2=0
(a-3)/x^2=1
(a-3)=x^2
(a-3)^1/2=x
now it is not possible to have a>3 because then x is complex.
so a<=3

Am I correct?
If Yes, how can I find relative extrema now?
Pls help me . Thanx.

 

[fcabanski]

There is a mistake.

a-3 >=0
a >=3.

 

[DrPhil]

question:- Find relative extremum of f(x)= x+(a-3)/x ;a€R and
Find the points where the function is increasing.

My try: f´(x)= 1-(a-3)/x^2
To find points f´(x)=0
1-(a-3)/x^2=0
(a-3)/x^2=1
(a-3)=x^2
(a-3)^1/2=x
now it is not possible to have a>3 because then x is complex. Wrong way round!
so a<=3

Am I correct?
If Yes, how can I find relative extrema now?
Pls help me . Thanx.

\(\displaystyle f(x)= x+(a-3)/x\)

\(\displaystyle \displaystyle f'(x) = 1 - (a - 3)/x^2\)

..........\(\displaystyle \displaystyle = \dfrac{x^2 -a + 3}{x^2}\)

By writing the two terms of the derivative over a common denominator, you can find 0s just by setting the numerator to 0. It is also "obvious" that neither the function nor the derivative are defined for \(\displaystyle x=0\).

......\(\displaystyle f'(x) = 0 \implies x^2 = a-3 \implies x = \pm\sqrt{a-3}\)

You can't specify \(\displaystyle a\), but you can make a statement about whether \(\displaystyle f'(x)\) has any 0s.
"If \(\displaystyle a \le 3\), there are no relative min or max of \(\displaystyle f(x)\)"

To find the relative extrema when \(\displaystyle a>3\), evaluate \(\displaystyle f(x)\) at the two roots of \(\displaystyle f'(x)\).

 

[fcabanski]

Isn't that the same as what the OP did, except for the mistake?

\(\displaystyle 1-\frac{a-3}{x^2}=0\)

\(\displaystyle 1=\frac{a-3}{x^2}\)

\(\displaystyle x^2=a-3\)

\(\displaystyle x = \pm \sqrt{a-3}\)

 

[DrPhil]

Isn't that the same as what the OP did, except for the mistake?

\(\displaystyle 1-\frac{a-3}{x^2}=0\)

\(\displaystyle 1=\frac{a-3}{x^2}\)

\(\displaystyle x^2=a-3\)

\(\displaystyle x = \pm \sqrt{a-3}\)

YES, navigator did it right (I hadn't seen the correction when I answered). Needed clarification of what happens if a<3, and still needs to find f(x) at the extreme points.